Malcev presentations for subsemigroups of direct products of ...

Let U be the subset of T where each padding string is either empty or oneletter long — elements of R are either adjacent or separated by a single (a, a)for some a ∈ A. Observe that the set U is finite because Y — which dictates howmany elements of R can appear — is finite. . Every relation in T is a Malcev consequence of those in U.Proof of 18. Let (αwβ, γwζ) be a relation in T, with (w, w) being padding. Suppose(w, w) = (a, a)(w ′ , w ′ ) for a ∈ A. Since0 H = (αwβ, γwζ)δ= (αwβ)ρ H − (γwζ)ρ H= (αβ)ρ H − (γζ)ρ H + (w)ρ H − (w)ρ H= (αβ)ρ H − (γζ)ρ H= (αβ, γζ)δ,the relations(αaβ, γaζ), (αw ′ β, γw ′ ζ), (αβ, γζ)()are all in ker ρ. They are all clearly in T.The following Malcev chain shows that (αwβ, γwζ) is a Malcev consequenceof the relations ():αwβ→ αaw ′ β→ αaββ R w ′ β→ γaζβ R w ′ β→ γaγ L γζβ R w ′ β→ γaγ L αββ R w ′ β→ γaγ L αw ′ β→ γaγ L γw ′ ζ→ γaw ′ ζ= γwζ.by induction on |w|Apply such reasoning to every padding string in the relation to show that it isa Malcev consequence of U. 18Conclusion. By Lemmata , , and ,ker ρ = S M = T M = U M .Therefore S admits the finite Malcev presentation SgM⟨A|U⟩. Since S was an arbitraryfinite generated subsemigroup G, the group G — which was an arbitrarydirect product of a free group and an abelian group — is Malcev coherent. 9 The direct product of a free group and a polycyclic group is coherent.Of course, Corollary implies that such direct products are not in generalMalcev coherent. This leaves the following question unanswered: