Malcev presentations for subsemigroups of direct products of ...

elation in ker ρ is a Malcev consequence of relations that have a decomposition() where each pair (c i , d i ) is drawn from a particular finite set. However, theset of such relations is manifestly infinite. The second — rather technical —stage involves showing that all these relations are Malcev consequences of thosein a different, but still infinite, set. The reader — although perhaps beginning toempathize with Sisyphus — should be reassured by the fairly simple structureof this new set of relations. The third stage is an easy proof that a finite subsetof these relations suffices for a Malcev presentation.Preliminaries. Let S F = Sπ F ⊆ F. Notice that S F is not in general a subset of Sand that Aρ F is a finite generating set for S F . LetJ(A) = {uv −1 : u, v ∈ A + , (u, v) ∈ ker ρ F }.Using the results of [MS], one can easily show that J(A) is a context-free languageover A∪A −1 . (See [CRRb] for the reasoning in full.) Let Γ be a contextfreegrammar recognizing J(A). For (u, v) ∈ ker ρ F , define n(u, v) to be theminimum number of internal vertices in a Γ-derivation tree for uv −1 ∈ J(A).(See [HU, § .] for information on derivation trees.) For the purposes of thisproof, a path from the root of a Γ-derivation tree to an external vertex (labelledby an element of A ∪ A −1 ) is called a derivation path.Let R be the subset of ker ρ F consisting of all relations (u, v) such that uv −1admits a Γ-derivation tree in which every derivation path contains at most twovertices labelled by the same non-terminal. The set R is finite. (Theorem of[CRRb] shows that subsemigroup S F has a finite Malcev presentation SgM⟨A|R⟩.)Assume R is symmetrized, so that (u, v) ∈ R implies that (v, u) ∈ R. SupposethatR = {(u 1 , v 1 ), (v 1 , u 1 ), . . . , (u n , v n ), (v n , u n )} ,where u i , v i ∈ A + . The set of relations R is contained in ker ρ F . Fix these pairs(u i , v i ) throughout the proof.Define δ : A + × A + → H by (u, v)δ = (uρ H ) − (vρ H ). Let D = Rδ ⊆ H.Observe that (u, v)δ = −(v, u)δ. Throughout this proof, δ is used as a measureof the ‘difference’ in the H-components of the elements represented by the twosides of a relation in ker ρ F .Each (u, v) ∈ ker ρ can be decomposed (possibly in many ways) as a product(u, v) = (c 1 , d 1 )(c 2 , d 2 ) · · · (c k , d k ),where u = c 1 c 2 · · · c k , v = d 1 d 2 · · · d k , and (c i , d i ) ∈ ker ρ F .Define, for each decomposition (c 1 , d 1 )(c 2 , d 2 ) · · · (c k , d k ),andn ′ ((c 1 , d 1 )(c 2 , d 2 ) · · · (c k , d k )) = max{n(c i , d i ) : i = 1, . . . , k}.p ′ ((c 1 , d 1 )(c 2 , d 2 ) · · · (c k , d k )) =∣ {i : n(ci , d i ) = n ′ ((c 1 , d 1 )(c 2 , d 2 ) · · · (c k , d k ))} ∣ ∣ .So n ′ is the maximum n-value of any of the (c i , d i ), and p ′ is the number oftimes this maximum is achieved.Fix a canonical decomposition of each relation (u, v) ∈ ker ρ by selecting thedecompositions that minimize n ′ and from these selecting one that minimizesp ′ . Definen ′′ (u, v) = n ′ ((c 1 , d 1 )(c 2 , d 2 ) · · · (c k , d k ))