On cancellative congruences for semigroups
Olga Macedońska · 1999
Given a semigroup identity u = v. We describe a smallest normal subgroup N in a free group F, such that F/N contains a relatively free cancellative semigroup which satisfies the identity u = v. Let F be a free group and F be a free semigroup (F 3 1), both generated by the same set M = {x, y, z, . . .}. A semigroup identity of a group G (or a semigroup S) is a nontrivial identity of the form u = v where u, v ∈ F , which holds under every substitution of generators by elements from G (elements from S). Research concerning semigroup identities was initiated by A.I. Mal’cev in early fifties and is still continued (see e.g. [10], [6], [18], [13] – [15], [11], [7]). One of the open questions concerning semigroup identities in a group G and a subsemigroup S generating G is whether an identity can hold in S without holding in G [1]. Partial answers to this question are given in [9], [5] and [2]. We give here some technical result which relates a cancellative congruence in F , providing an identity in a quotient semigroup, and a normal subgroup it defines in F. A congruence in F , providing the identity u = v has to contain (u, v) and the set which is the smallest invariant (under endomorphisms of F) reflexive, and symmetric closure of (u, v). We denote this set by (u, v). To extend (u, v) to a transitive relation we define a step. Two words a, b ∈ F are called connected by (u, v)-step if a = c1sc2, b = c1tc2, and (s, t) ∈ (u, v). It is clear that if (s, t) ∈ (u, v), then (a, b) ∈ (u, v). However if a = c1sc2 → c1tc2 = d1sd2 → d1td2 = b, where each arrow denotes (u, v)-step, then it is not necessary that (a, b) ∈ (u, v). A congruence in F , providing the identity u = v in the quotient semigroup without necessity to be cancellative is described in [4]. Namely, two words are congruent if and only if they are connected by a finite sequence of (u, v)-steps. AMS subject classification: Primary 20E10; Secondary 20M07.