Repnitskii proved that any lattice embeds in a subsemigroup lattice of some commutative, cancellative, idempotent free semigroup with unique roots. In that proof, use is made of a result by Bredikhin and Schein stating that any lattice embeds in a suborder lattice of suitable partial order. Here, we present a direct proof of Repnitskii's result which is independent of Bredikhin-Schein's, thus giving the answer to the question posed by Shevrin and Ovsyannikov.
PRELIMINARIESFor an algebraic structure A, Sub(A) denotes the set of all substructures of A. Ordered by inclusion, Sub(A) forms a complete algebraic lattice, wheresubstructures in A. (Here, for any X ⊆ A, X denotes the substructure of A generated by X.) Following [1], we say that a class U of algebraic structures (of a fixed signature) is lattice universal if any lattice embeds in Sub(A), for some A ∈ U. At present we know of several examples of lattice-universal classes of structures. Among these are the class of all groups [2], the class of partially ordered sets [3; see also 4], and various classes of semigroups [5]. The results on lattice-universal classes of semigroups are summarized in THEOREM 1.1 [1, Thm. 27.2]. The property of being lattice universal is shared by the following classes of semigroups: (i) the class of all commutative nil semigroups of index 2; (ii) the class of all semilattices;(iii) the class of commutative, cancellative, idempotent free semigroups with unique roots. The original proof of this result in [5] made essential use of the above-mentioned result that any lattice embeds in a suborder lattice of suitable partial order. Namely, in [5], it was proved that any suborder *