Malcev presentations for subsemigroups of direct products of ...

Malcev presentations for subsemigroupsof direct products of coherent groupsAlan J. CainSchool of Mathematics and Statistics, University of St AndrewsNorth Haugh, St Andrews, Fife KY SS, United KingdomEmail: alanc@mcs.st-andrews.ac.ukWeb page: www-groups.mcs.st-andrews.ac.uk/~alanc/This is a preprint of the article published as: Journal of Pure and AppliedAlgebra (), no. , pp. –. doi: ./j.jpaa....: . : ..The direct product of a free group and a polycyclic group is knownto be coherent. This paper shows that every finitely generatedsubsemigroup of the direct product of a virtually free group andan abelian group admits a finite Malcev presentation. (A Malcevpresentation is a presentation of a special type for a semigroupthat embeds into a group. A group is virtually free if it containsa free subgroup of finite index.) By considering the direct productof two free semigroups, it is also shown that polycyclic groups,unlike nilpotent groups, can contain finitely generated subsemigroupsthat do not admit finite Malcev presentations. Malcev presentations are presentations of a special type for semigroupsembeddable into groups. They were introduced by Spehner [Spe],who later proved that all finitely generated submonoids of free monoids admitfinite Malcev presentations [Spe]. These two articles by Spehner formedthe whole of the literature on Malcev presentations until recent work of Cain,Robertson & Ruškuc [CRRa, CRRb] and Cain [Cai]. The present paper continuesthis research into Malcev presentations.Call a group — or more generally a group-embeddable semigroup — Malcevcoherent if all of its finitely generated subsemigroups admit finite Malcevpresentations. (Recall that a group is coherent if all of its finitely generated subgroupsare finitely presented [Ser].) The class of Malcev coherent groups isproperly contained in the class of coherent groups [CRRa, Theorem ].The direct product of two free non-abelian groups is not coherent [Gru].This does not immediately preclude the Malcev coherence of the direct product

of two free semigroups, but Theorem shows that such direct products are notin general Malcev coherent: the direct product of two free semigroups of rankat least 2 is not Malcev coherent.The Malcev coherence of nilpotent groups was established by [CRRa, Theorem]. Polycyclic groups form a more general class than finitely generatednilpotent groups whilst retaining the property of coherence [Sim, §§ .–.].One therefore naturally asks whether polycyclic groups are Malcev coherent.From Theorem and a result of Rosenblatt [Ros], one obtains a negative answer:polycyclic groups are not in general Malcev coherent.The direct product of a free group and an abelian group is coherent. (Aslight modification of the reasoning of [Mil] proves this assertion and, moregenerally, establishes the coherence of the direct product of a free group and apolycyclic group.) Both virtually free groups [CRRb, Theorem ] and abeliangroups are known to be Malcev coherent. [Every finitely generated subsemigroupof an abelian group — and more generally every finitely generated commutativesemigroup — admits a finite ‘ordinary’ presentation as a consequenceof Rédei’s Theorem [Réd].] Most of this paper is dedicated to the proof of itssecond main result, Theorem , which asserts that every direct product of avirtually free group and an abelian group is Malcev coherent.[This paper is based on Chapter of the author’s Ph.D. thesis [Cai].] Following [ECH + ], the notation used in this paper distinguishesa word from the element of the semigroup or group it represents. Let A be analphabet representing a set of generators for a semigroup S. For any word w ∈A + , denote by w the element of S represented by w. Similarly, if A representsa set of generators for a group G, let w be the element of G represented byw ∈ (A ∪ A −1 ) ∗ . In both cases, for any set of words W, W is the set of allelements represented by at least one word in W.The theory of Malcev presentations was introduced by Spehner [Spe], thoughthey are based on Malcev’s necessary and sufficient condition for the embeddabilityof a semigroup in a group [Mal, Mal]. (Details of the embeddabilitycondition can also be found in [CP, Chapter ].) Although this sectioncontains the definitions and results about Malcev presentations required for therest of the paper, the reader is assumed to be familiar with the basic theory of[ordinary] semigroup presentations. For a fuller exposition of the foundationsof the theory of Malcev presentations, see [Cai, Chapter ]. . Let S be any semigroup. A congruence σ on S is a Malcevcongruence if S/σ is embeddable in a group.If {σ i : i ∈ I} is a set of Malcev congruences on S, then σ = ∩ i∈I σ i is alsoa Malcev congruence on S. This is true because S/σ i embeds in a group G i foreach i ∈ I, so S/σ embeds in ∏ i∈I S/σ i, which in turn embeds in ∏ i∈I G i. Thefollowing definition therefore makes sense. . Let A + be a free semigroup; let ρ ⊆ A + × A + be any binaryrelation on A + . Denote by ρ M the smallest Malcev congruence containing ρ —namely,ρ M = ∩ {σ : σ ⊇ ρ, σ is a Malcev congruence on A+ } .

Then SgM⟨A | ρ⟩ is a Malcev presentation for [any semigroup isomorphic to]A + /ρ M . If both A and ρ are finite, the the Malcev presentation SgM⟨A | ρ⟩ issaid to be finite.Let SgM⟨A | R⟩ be a Malcev presentation for a semigroup S. If S has a finiteMalcev presentation, then it admits one of the form SgM⟨A | Q⟩, where Q is afinite subset of R.The notation SgM⟨A | ρ⟩ distinguishes the Malcev presentation with generatorsA and defining relations ρ from the ordinary semigroup presentationSg⟨A | ρ⟩, which defines A + /ρ # . (Recall that ρ # denotes the smallest congruencecontaining ρ.) Similarly, Gp⟨A|ρ⟩ denotes the group presentation with the sameset of generators and defining relations. Let X be a subset of a group G. Denoteby Sg⟨X⟩ the subsemigroup generated by X, by Mon⟨X⟩ the submonoid generatedby X, and by Gp⟨X⟩ the subgroup generated by X. The free group withbasis A is denoted FG(A).Fix A + and ρ as in the Definition and let S = A + /ρ M . Let A L , A R be twosets in bijection with A under the mappings a ↦→ a L , a ↦→ a R , respectively, withA, A L , A R being pairwise disjoint. Extend the mappings a ↦→ a L , a ↦→ a R toanti-isomorphisms from A ∗ to (A L ) ∗ and (A R ) ∗ , respectively. Letτ = ρ ∪ { (bb R a, a), (abb R , a), (b L ba, a), (ab L b, a) : a ∈ A ∪ A L ∪ A R , b ∈ A } .Let G = Sg⟨A ∪ A L ∪ A R | τ⟩. The semigroup G is actually the universal groupof S, with G ≃ Gp⟨A | ρ⟩. (See [CP, Chapter ] for background informationon universal groups.) Therefore a Malcev presentation for a semigroup is effectivelya presentation for its universal group, with all defining relations confinedto A + × A + . It can be shown thator equivalently thatρ M = τ # ∩ (A + × A + ),ρ M = {(u, v) : u, v ∈ A + represent the same element of Gp⟨A | ρ⟩}.()Therefore, two words u, v ∈ A + represent the same element of S if and only ifthere is a sequenceu = u 0 → u 1 → . . . → u n = v,with n ⩾ 0, where, for each i ∈ {0, . . . , n − 1}, there exist p i , q i , q ′ i , r i ∈ (A ∪A L ∪ A R ) ∗ such that u i = p i q i r i , u i+1 = p i q ′ i r i, and (q i , q ′ i ) ∈ τ or (q′ i , q i) ∈ τ.In fact, it can be shown that u, v ∈ A + represent the same element of S if anyonly if there exists such a sequence with p i ∈ (A ∪ A L ) ∗ and r i ∈ (A ∪ A R ) ∗ .This restriction on the letters that can appear in p i and r i simply means that nochanges can occur to the left of an a L or to the right of an a R . Such a sequenceis called a Malcev ρ-chain (or simply a Malcev chain) from u to v. If (u, v) ∈ ρ M ,then (u, v) is said to be a Malcev consequence of ρ. The first part of the present section is devoted to a proof that thedirect product of two free semigroups of rank at least 2 is not Malcev coherent.This contrasts with Spehner’s proof of the Malcev coherent of free monoids

[Spe]. This in turn contrasts the fact that free semigroups admit finitely generatedsubsemigroups that do not admit finite ‘ordinary’ presentations. [However,all subsemigroups of rank at most 3 of free semigroups do admit finite ordinarypresentations [BMa]. Rank-4 subsemigroups of free semigroups maynot be finitely presented [Mar, p.].]The following example exhibits a finitely generated subsemigroup of a directproduct of two free semigroups that does not admit a finite Malcev presentation. . Let A = {a, b, c, d, e, f, g, h, i, j} be an alphabet representing elementsof {x, y, p, q, r, s} + × {x, y, p, q, r, s} + as follows:a = (x 2 pqrs, x),b = (pqrspqrs, p),c = (pqr, q),d = (spqr, r),e = (sy 2 , s),f = (x 2 pq, x),g = (rspq, p),h = (rsp, q),i = (qrspqrsp, r),j = (qrsy 2 , s).Let S be the subsemigroup of {x, y, p, q, r, s} + ×{x, y, p, q, r, s} + generated by A. . The semigroup S is presented by Sg⟨A | R⟩, whereR = {(ab α cd α e, fg α hi α j) : α ∈ N ∪ {0}} .[Observe in passing that (A, R) is a complete (i.e. confluent and noetherian)rewriting system. This fact is not essential for any of the subsequent reasoningin the paper.]Proof of 4. Every relation in R holds in S:ab α cd α e = (x 2 pqrs(pqrspqrs) α pqr(spqr) α sy 2 , xp α qr α s)= (x 2 (pqrs) 3α+2 y 2 , xp α qr α s)= (x 2 pq(rspq) α rsp(qrspqrsp) α qrsy 2 , xp α qr α s)= fg α hi α jfor all α ∈ N ∪ {0}.Define a set N of normal forms to be the set of all words in A + that do notcontain subwords of the form fg α hi α j for any α ∈ N ∪ {0}. Clearly, since thereare no overlaps between these forbidden words and words ab α cd α e, every elementof S is represented by at least one element of N.Let u = u 1 · · · u ξ and v = v 1 · · · v η (u i , v i ∈ A for all i) be distinct words inthe set of normal forms N, and suppose they represent the same element of S.Without loss of generality, suppose that u 1 ≠ v 1 and that u precedes v in thelexicographic ordering based on a ≺ b ≺ c ≺ . . . ≺ j. Consider the secondcomponent of u 1 and v 1 to see that(u 1 , v 1 ) ∈ {(a, f), (b, g), (c, h), (d, i), (e, j)}.Examining the first component forces u 1 = a and v 1 = f. Sou = (x 2 pqrs · · · , x · · · ),v = (x 2 pq · · · , x · · · ).

The first component of v 2 must begin with r, which shows that the letter v 2 iseither g or h, and consideration of second components then forces u 2 = b oru 2 = c, respectively. Without loss of generality, assume that the next α ∈ N∪{0}letters of v are letters g and v α+2 ≠ g. The case α > 0 is explained fully below;the case α = 0 is similar.So suppose v 2 · · · v α+1 = g α and v α+2 ≠ g with α > 0. Thenor, rearranging parentheses,u = (x 2 pqrs · · · , x · · · ),v = (x 2 pq(rspq) α · · · , xp α · · · ).u = (x 2 pqrs · · · , x · · · ),v = (x 2 (pqrs) α pq · · · , xp α · · · ).[In the remainder of the present proof, parentheses will be rearranged withoutcomment.] In order to match the p α in the second component of v, the lettersu 2 , …u α+1 all lie in {b, g}. Since the first component of u 2 must begin with p(because α > 0), u 2 = b. So u begins ab β u β+2 · · · , where u β+2 ≠ b, withβ > 0.Suppose β < α/2. Then the first component of u β+2 must begin with p,and so u β+2 = c. The second component of u is then xp β q · · · and that of v isxp α · · · . However, this contradicts the equality of u and v since α > 2β > β.Therefore β ⩾ α/2 and the first component of v α+2 begins with rs, whencev α+2 = h (since v α+2 ≠ g). Sou = (x 2 (pqrs) 2β+1 · · · , xp β · · · ),v = (x 2 (pqrs) α+1 p · · · , xp α q · · · ).This forces β ⩽ α, and either u β+2 = c or u β+2 = h.Suppose β = α/2. Then the first component of u β+2 begins with p and sou β+2 = c. Consideration of the second components of u and v again leads to acontradiction.Therefore β > α/2. Assume that 2β = α + γ, where γ > 0. Sou = (x 2 (pqrs) α+γ+1 · · · , xp (α+γ)/2 · · · ),v = (x 2 (pqrs) α+1 p · · · , xp α q · · · ).Consideration of first components requires that all of the letters v α+3 , …, v α+δ+2must be i, where δ ⩾ ⌊γ/2⌋. Assume without loss of generality that v α+δ+3 ≠ i.Thereforeu = (x 2 (pqrs) α+γ+1 · · · , xp (α+γ)/2 · · · ),v = (x 2 (pqrs) α+2δ+1 p · · · , xp α qr δ · · · ).Suppose first that γ is odd and that δ = (γ − 1)/2.Then the first componentof v α+δ+3 would begin with qrs, forcing v α+δ+3 = j. Thenu = (x 2 (pqrs) α+γ+1 · · · , xp (α+γ)/2 · · · ),v = (x 2 (pqrs) α+γ+1 y 2 · · · , xp α qr γ−1 s · · · ).So the first component of u β+2 begins with y 2 , which is a contradiction.

Therefore δ > (γ − 1)/2. Assume that 2δ = γ − 1 + ϵ, where ϵ > 0. Then thefirst component of u β+2 begins with p, forcing u β+2 = c. Thenu = (x 2 (pqrs) α+γ+1 pqr · · · , xp (α+γ)/2 q · · · ),v = (x 2 (pqrs) α+γ+ϵ p · · · , xp α qr δ · · · ).Suppose ϵ = 1. Since v α+δ+3 begins with q and v α+δ+3 ≠ i, it must holdthat v α+δ+3 = j, which forces u β+3 = e. Considering second componentsthen requires δ = 0, which is impossible. So ϵ ⩾ 2 and the first component ofu β+3 · · · u ξ must begin (spqr) ϵ−2 sp. So the letter u β+3 must begin a string ofϵ − 1 letters d. That is, u β+3 , …, u β+ϵ+1 must all be d. This givesu = (x 2 (pqrs) α+γ+ϵ pqr · · · , xp (α+γ)/2 qr ϵ−1 · · · ),v = (x 2 (pqrs) α+γ+ϵ p · · · , xp α qr δ · · · ).The first component of v α+δ+3 must begin qr, which forces v α+δ+3 = j. Thefirst component of u β+ϵ+2 must begin sy 2 , which forces u β+ϵ+2 = e. Sou = (x 2 (pqrs) α+γ+ϵ+1 y 2 · · · , xp (α+γ)/2 qr ϵ−1 s · · · ),v = (x 2 (pqrs) α+γ+ϵ+1 y 2 · · · , xp α qr δ s · · · ).Examine the second components of u and v to get (α+γ)/2 = α and ϵ−1 = δ.Since 2δ = γ − 1 + ϵ, it follows that α = γ = δ. Sov = v 1 (v 2 · · · v α+1 )v α+2 (v α+3 · · · v α+δ+2 )v α+δ+3 · · ·= fg α hi α j · · · ,which contradicts v’s membership of the set of normal forms N. 4The semigroup S is therefore isomorphic to that of [CRRa, § ], which doesnot admit a finite Malcev presentation. To see this, observe that the universalgroup of S iswhereG = Gp⟨A | R⟩ ≃ FG(a, b, c, d, e) ∗ K FG(f, g, h, i, j) ,K = Gp⟨ab α cd α e : α ∈ N ∪ {0}⟩ ≃ Gp⟨fg α hi α j : α ∈ N ∪ {0}⟩ .The generating set {ab α cd α e : α ∈ N ∪ {0}} forms a basis for the amalgamatedsubgroup K by [LS, Proposition I..]. So K is not finitely generated, and soa theorem of Baumslag [Bau] shows that the free product G is not finitelypresented. Thus, as its universal group is not finitely presented, S cannot admita finite Malcev presentation. . The direct product of two free semigroups of rank at least 2 is not Malcevcoherent.Proof of 5. Let D be the direct product of two free semigroups of rank at least2. The free semigroup of rank 2 contains isomorphic copies of free semigroupsof every rank; D therefore contains a subsemigroup isomorphic to the directproduct {x, y, p, q, r, s} + × {x, y, p, q, r, s} + . The semigroup D thus contains thefinitely generated subsemigroup S of Example , which does not admit a finiteMalcev presentation. Therefore D is not Malcev coherent. 5

Every finitely generated nilpotent group is polycyclic [Sim, Proposition ..].Whilst the class of polycyclic groups is strictly larger than the class of finitelygenerated nilpotent groups, the former class retains many of the pleasant propertiesof the latter [Sim, § .]. In particular, polycyclic groups are coherent.This remainder of this section is dedicated to proving that polycyclic groups arenot in general Malcev coherent, which contrasts with the Malcev coherence ofvirtually nilpotent groups [CRRa, Theorem ].The following two results are needed: ([Sim, Proposition ..]). Every extension of a polycyclic groupby a polycyclic group if itself polycyclic. That is, if E is an extension of G, and G andE/G are both polycyclic, then E is polycyclic. In particular, the direct product of twopolycyclic groups is polycyclic. ([Ros, Theorem .]). Let G be a polycyclic group. Then exactlyone of the following two statements is true:1. The group G is virtually nilpotent.2. The group G contains a free subsemigroup of rank 2.The fact that polycyclic groups are not in general Malcev coherent followsas a consequence of these results and Theorem : . The direct product of two polycyclic groups that are not virtuallynilpotent is not Malcev coherent. Therefore polycyclic groups are not in general Malcevcoherent.Proof of 8. Let G and H be polycyclic groups that are not virtually nilpotent. LetP = G × H. Theorem shows that G and H both contain a free subsemigroupof rank 2. Therefore P contains the direct product of two free semigroups ofrank 2, which is not Malcev coherent by Theorem . Ergo, P itself is not Malcevcoherent. By Proposition , P is a polycyclic group. 8 . Every direct product of a virtually free group and an abelian group isMalcev coherent.Proof of 9. Let F be a virtually free group and let H be an abelian group. LetA be a finite alphabet representing elements of G = F × H. Let ρ : A ∗ → Gbe the standard representation mapping. Let S = Sg⟨Aρ⟩. The semigroup S isobviously presented by Sg⟨A | ker ρ⟩.Let π F : G → F and π H : G → H be the projection mappings to F and H,respectively. Define ρ F : A ∗ → F and ρ H : A ∗ → H by ρπ F and ρπ H , respectively.Notice that ker ρ = ker ρ F ∩ ker ρ H .The strategy of the proof is based on the observation that any relation (u, v) ∈ker ρ can be decomposed as(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 each (c i , d i ) is in ker ρ F but notnecessarily in ker ρ H . The first stage of the proof involves showing that every

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 ))

andp ′′ (u, v) = p ′ ((c 1 , d 1 )(c 2 , d 2 ) · · · (c k , d k )),where (c 1 , d 1 )(c 2 , d 2 ) · · · (c k , d k ) is the canonical decomposition of (u, v).First stage. Let S be the subset of ker ρ consisting of those relations whose canonicaldecompositions are formed by concatenating elements of R. LetQ = {(u i wv i , v i wu i ) : w ∈ A ∗ and i = 1, . . . , n}.Notice that Q ∈ R # , and furthermore that if (p, q) ∈ R + and w ∈ A ∗ , then(pwq, qwp) is a consequence of Q. (The set R + consists of all relations formedby concatenating elements of R.)Define an ordering ≪ of the set ker ρ as follows:(u, v) ≪ (u ′ , v ′ ) ⇐⇒ n ′′ (u, v) < n ′′ (u ′ , v ′ ) or(n ′′ (u, v) = n ′′ (u ′ , v ′ ) and p ′′ (u, v) < p ′′ (u ′ , v ′ ) ) .The line of reasoning in this first stage owes much to the proof of [CRRb,Theorem ]. To show that each (u, v) is a Malcev consequence of ≪-precedingelements of ker ρ, one follows the basic outline of that earlier proof to obtain ≪-preceding elements of ker ρ F ; one ‘compensates’ for the fact that these relationsmay not lie in ker ρ H by inserting pairs u i u R i , v iv R i , uL i u i, or v L i v i; and one usesthese newly-found relations and those in Q in a Malcev chain yielding (u, v). . Let (u, v) ∈ ker ρ − S. Then (u, v) is a Malcev consequence of relationsin Q and ≪-preceding elements of ker ρ.The following technical result will be needed in the proof of Theorem .Informally, it is this result that allows the ‘compensation’ mentioned above. . Let (u, v) ∈ ker ρ F . Then (u, v)δ is a positive (that is, semigroup) sumof elements of D.Proof of 11. This result is obviously true for elements of R ⊆ ker ρ F . Proceedby induction on n(u, v). Suppose (u, v) ∈ ker ρ F − R. Then any Γ-derivationtree fro uv −1 contains a derivation path with at least three vertices labelled bythe same non-terminal. Distinguish such a path with m > 2 repetitions of thenon-terminal M. SupposeO ∗ ⇒ xMy, M ∗ ⇒ α i Mβ i for each i ∈ {1, . . . , m − 1}, M ∗ ⇒ w,where O is the start symbol of Γ and uv −1 = xα 1 · · · α m−1 wβ m−1 · · · β 1 y.The point in the word uv −1 where the subword u ∈ A + ends and the subwordv −1 ∈ (A −1 ) + begins is called the u–v −1 boundary. . The u–v −1 boundary is in either x, w, or y (possibly at the end of x orw or the start of w or y).Proof of 12. Suppose the u–v −1 boundary is in α i for some i ∈ {1, . . . , m − 1}(not at the start of α 1 or the end of α m−1 ). Then there exist s, t ∈ A + such thatα 1 · · · α m−1 = st −1 . By pumping derivation from M, it can be seen thatJ(A) ∋ x(α 1 · · · α m−1 ) 2 w(β m−1 · · · β 1 ) 2 y= xst −1 st −1 wβ m−1 · · · β 1 β m−1 · · · β 1 y,which is a contradiction, because this word is not in A + (A −1 ) + . A similar contradictionarises should the u–v −1 boundary be in some β i , thus proving thelemma. 12

The relation (u, v) therefore takes one of the following three forms:. (xα 1 · · · α m−1 w ′ , y −1 β −11 · · · β−1 m−1 w′′ ), where w = w ′ (w ′′ ) −1 , if the u–v −1boundary is in w;. (xα 1 · · · α m−1 wβ m−1 · · · β 1 y ′ , y ′′ ), where y = y ′ (y ′′ ) −1 , if the u–v −1 boundaryis in y;. (x ′ , y −1 β −11 · · · β−1 m−1 w−1 α −1m−1 · · · α−1 1 x′′ ), where x = x ′ (x ′′ ) −1 , if the u–v −1 boundary is in x.The second and third cases are symmetrical. It shall therefore suffice to provethe result for the first two cases.Observe that in Γ, since m > 2, O ∗ ⇒ xMy, M ∗ ⇒ α 1 · · · α m−2 Mβ m−2 · · · β 1 ,M ∗ ⇒ α m−1 Mβ m−1 and M ∗ ⇒ w, and thereforexα 1 · · · α m−2 wβ m−2 · · · β 1 y,xα m−1 wβ m−1 y, xwy ∈ J(A).()Furthermore, derivation trees with fewer than n(u, v) internal vertices exist foreach of these words, as the following three derivations show:O ∗ ⇒ xMy ∗ ⇒ xα 1 · · · α m−2 Mβ m−2 · · · β 1 y ∗ ⇒ xα 1 · · · α m−2 wβ m−2 · · · β 1 y,O ∗ ⇒ xMy ∗ ⇒ xα m−1 Mβ m−1 y ∗ ⇒ xα m−1 wβ m−1 y,O ∗ ⇒ xMy ∗ ⇒ xwy.. Supposeu = xα 1 · · · α m−1 w ′ and v = y −1 β −11 · · · β−1 m−1 w′′ .Then, by (), the relations(xα 1 · · · α m−2 w ′ , y −1 β −11 · · · β−1 m−2 w′′ ),(xα m−1 w ′ , y −1 β −1m−1 w′′ ), and (xw ′ , y −1 w ′′ )are in ker ρ F and have n-values less than n(u, v). Assume that δ applied toeach of these relations () gives a positive sum of elements of D. Now,(u, v)δ = (uρ H ) − (vρ H )= (xα 1 · · · α m−1 w ′ )ρ H − (y −1 β −11 · · · β−1 m−1 w′′ )ρ H= (xα 1 · · · α m−2 w ′ )ρ H + (α m−1 )ρ H− (y −1 β −11 · · · β−1 m−2 w′′ )ρ H − (β m−1 )ρ H= (xα 1 · · · α m−2 w ′ , y −1 β −11 · · · β−1 m−2 w′′ )δ + (α m−1 )ρ H − (β m−1 )ρ H= (xα 1 · · · α m−2 w ′ , y −1 β −11 · · · β−1 m−2 w′′ )δ + (α m−1 )ρ H − (β m−1 )ρ H+ (x)ρ H + (w ′ )ρ H − (y −1 )ρ H − (w ′′ )ρ H− (x)ρ H − (w ′ )ρ H + (y −1 )ρ H + (w ′′ )ρ H= (xα 1 · · · α m−2 w ′ , y −1 β −11 · · · β−1 m−2 w′′ )δ+ (xα m−1 w ′ )ρ H − (y −1 β m−1 w ′′ )ρ H− (xw ′ )ρ H + (y −1 w ′′ )ρ H= (xα 1 · · · α m−2 w ′ , y −1 β −11 · · · β−1 m−2 w′′ )δ+ (xα m−1 w ′ , y −1 β m−1 w ′′ )δ + (y −1 w ′′ , xw ′ )δ.The assumption then shows that (u, v)δ is a positive sum of elements of D.()

. SupposeThen, by (), the relationsu = xα 1 · · · α m−1 wβ m−1 · · · β 1 y ′ and v = y ′′ .(xα 1 · · · α m−2 wβ m−2 · · · β 1 y ′ , y ′′ ),(xα m−1 wβ m−1 y ′ , y ′′ ), and (xwy ′ , y ′′ )are in ker ρ F and have n-values less than n(u, v). Again assume that δ appliedto each of these relations () gives a positive sum of elements of D.Then(u, v)δ = (uρ H ) − (vρ H )= (xα 1 · · · α m−1 wβ m−1 · · · β 1 y ′ )ρ H − (y ′′ )ρ H= (xα 1 · · · α m−2 wβ m−2 · · · β 1 y ′ )ρ H − (y ′′ )ρ H+ (α m−1 )ρ H + (β m−1 )ρ H= (xα 1 · · · α m−2 wβ m−2 · · · β 1 y ′ , y ′′ )δ + (α m−1 )ρ H + (β m−1 )ρ H+ (xwy ′ )ρ H − (y ′′ )ρ H − (xwy ′ )ρ H + (y ′′ )ρ H= (xα 1 · · · α m−2 wβ m−2 · · · β 1 y ′ , y ′′ )δ+ (xα m−1 wβ m−1 y ′ )ρ H − (y ′′ )ρ H − (xwy ′ )ρ H + (y ′′ )ρ H= (xα 1 · · · α m−2 wβ m−2 · · · β 1 y ′ , y ′′ )δ+ (xα m−1 wβ m−1 y ′ , y ′′ )δ + (y ′′ , xwy ′ )δ.Again, the assumption shows that(u, v)δ is a positive sum of elements of D.Therefore, by induction on n(u, v), the image under δ of each (u, v) ∈ ker ρ Fcan be expressed as a positive sum of elements of D. 114.2. Let (c 1 , d 1 ) · · · (c k , d k ) be the canonical decomposition of (u, v). Since (u, v) ∈ker ρ − S, there exists (c j , d j ) ∈ ker ρ F − R. Reasoning as in the proof ofLemma , there are three cases, two of which are parallel. For brevity, lets = c 1 · · · c j−1 , t = c j+1 · · · c k , s ′ = d 1 · · · d j−1 , and t ′ = d j+1 · · · d k .. Supposec j = xα 1 · · · α m−1 w ′ and d j = y −1 β −11 · · · β−1 m−1 w′′where x, α i , w ′ , w ′′ , β i , y as in the proof of Lemma . Then the relations(xα 1 · · · α m−2 w ′ , y −1 β −11 · · · β−1 m−2 w′′ ),are in ker ρ F . The relations(sxα 1 · · · α m−2 w ′ t, s ′ y −1 β −11 · · · β−1 m−2 w′′ t ′ ),(xα m−1 w ′ , y −1 β −1m−1 w′′ ), and (xw ′ , y −1 w ′′ )(sxα m−1 w ′ t, s ′ y −1 β −1m−1 w′′ t ′ ), and (sxw ′ , s ′ y −1 w ′′ t ′ )are thus also in ker ρ F . Now, since (u, v)δ = 0 H ,(sxα 1 · · · α m−2 w ′ t, s ′ y −1 β −11 · · · β−1 m−2 w′′ t ′ )δ = −(α m−1 )ρ H + (β −1m−1 )ρ H,(sxα m−1 w ′ t, s ′ y −1 β −1m−1 w′′ t ′ )δ = −(α 1 · · · α m−2 )ρ H+ (β −1m−2 · · · β−1 1 )ρ H,(sxw ′ t, s ′ y −1 w ′′ t ′ )δ = −(α 1 · · · α m−1 )ρ H+ (β −1m−1 · · · β−1 1 )ρ H.()

Lemma asserts that there are positive sums of elements of D thatequal the left-hand sides of the above equations. That is, one can chooseelements of R, concatenate them, and get a relation in ker ρ F whose imageunder δ takes one of these three values. By switching the two sides of sucha relation, one can invert the image under δ. Therefore choose (p, q) and(p ′ , q ′ ) in R + such that (p, q)δ = (α m−1 )ρ H − (β −1m−1 )ρ H and (p ′ , q ′ )δ =(α 1 · · · α m−2 )ρ H −(β −1m−2 · · · β−1 1 )ρ H. Observe that the relations (pwq, qwp)and (p ′ wq ′ , q ′ wp ′ ) are consequences of Q, and that(p ′ p, q ′ q)δ = (p, q)δ + (p ′ , q ′ )δ = (α 1 · · · α m−1 )ρ H − (β −1m−1 · · · β−1 1 )ρ H.Using these relations (p, q) and (p ′ , q ′ ) as ‘compensation’, one obtainsthe relations(psxα 1 · · · α m−2 w ′ t, qs ′ y −1 β −11 · · · β−1 m−2 w′′ t ′ ),(sxα m−1 w ′ tp ′ , s ′ y −1 β −1m−1 w′′ t ′ q ′ ), and (q ′ qsxw ′ , p ′ ps ′ y −1 w ′′ t ′ ),()which lie in ker ρ since their images under δ are 0 H . By the choice of (p, q)and (p ′ , q ′ ), the relations () precede (u, v) in the ≪-ordering: for example,the decomposition of the first relation(p, q)(s, s ′ )(xα 1 · · · α m−2 w ′ , y −1 β −11 · · · β−1 m−2 w′′ )(t, t ′ )has a lesser value of n ′′ , or the same n ′′ -value and a smaller p ′′ -value, thanthe canonical decomposition of (u, v), so certainly the canonical decompositionof the first relation must have the same property.The following Malcev chain shows that (u, v) is a Malcev consequenceof the relations () and those in Q:sxα 1 · · · α m−1 w ′ t→ p L psxα 1 · · · α m−2 w ′ tt R (w ′ ) R α m−1 w ′ t→ p L qs ′ y −1 β −11 · · · β−1 m−2 w′′ t ′ t R (w ′ ) R α m−1 w ′ tby ()→ p L qs ′ y −1 β −11 · · · β−1 m−2 (y−1 ) L (s ′ ) L q L (q ′ ) L q ′ qs ′ y −1 w ′′ t ′ t R (w ′ ) R α m−1 w ′ t→ p L qs ′ y −1 β −11 · · · β−1 m−2 (y−1 ) L (s ′ ) L q L (q ′ ) L p ′ psxw ′ tt R (w ′ ) R α m−1 w ′ t by ()→ p L qs ′ y −1 β −11 · · · β−1 m−2 (y−1 ) L (s ′ ) L q L (q ′ ) L p ′ psxα m−1 w ′ t→ p L qs ′ y −1 β −11 · · · β−1 m−2 (y−1 ) L (s ′ ) L q L (q ′ ) L p ′ psxα m−1 w ′ tq ′ (q ′ ) R→ p L qs ′ y −1 β −11 · · · β−1 m−2 (y−1 ) L (s ′ ) L q L (q ′ ) L q ′ psxα m−1 w ′ tp ′ (q ′ ) R by Q #→ p L qs ′ y −1 β −11 · · · β−1 m−2 (y−1 ) L (s ′ ) L q L psxα m−1 w ′ tp ′ (q ′ ) R→ p L qs ′ y −1 β −11 · · · β−1 m−2 (y−1 ) L (s ′ ) L q L ps ′ y −1 β −1m−1 w′′ t ′ q ′ (q ′ ) Rby ()→ p L qs ′ y −1 β −11 · · · β−1 m−2 (y−1 ) L (s ′ ) L q L ps ′ y −1 β −1m−1 w′′ t ′→ p L qs ′ y −1 β −11 · · · β−1 m−2 (y−1 ) L (s ′ ) L q L ps ′ y −1 β −1m−1 w′′ t ′ qq R→ p L qs ′ y −1 β −11 · · · β−1 m−2 (y−1 ) L (s ′ ) L q L qs ′ y −1 β −1m−1 w′′ t ′ pq R by Q #→ p L qs ′ y −1 β −11 · · · β−1 m−1 w′′ t ′ pq R→ p L ps ′ y −1 β1 −1 · · · β−1 m−1 w′′ t ′ qq R→ s ′ y −1 β −11 · · · β−1 m−1 w′′ t ′ .. Supposec j = xα 1 · · · α m−1 wβ m−1 · · · β 1 y ′ and d j = y ′′

where x, α i , w, β i , y ′ , y ′′ are as in the proof of Lemma . Then the relations(xα 1 · · · α m−2 wβ m−2 · · · β 1 y ′ , y ′′ ),(xα m−1 wβ m−1 y ′ , y ′′ ), and (xwy ′ , y ′′ )()are in ker ρ F . Therefore the relations:(sxα 1 · · · α m−2 wβ m−2 · · · β 1 y ′ t, s ′ y ′′ t ′ ), (sxα m−1 wβ m−1 y ′ t, s ′ y ′′ t ′ ),are also in ker ρ F . Since (u, v)δ = 0 H ,and (sxwy ′ t, s ′ y ′′ t ′ )(sxα 1 · · · α m−2 wβ m−2 · · · β 1 y ′ t, s ′ y ′′ t ′ )δ = −(α m−1 )ρ H − (β m−1 )ρ H ,(sxα m−1 wβ m−1 y ′ t, s ′ y ′′ t ′ )δ = −(α 1 · · · α m−2 )ρ H − (β m−2 · · · β 1 )ρ H ,(sxwy ′ t, s ′ y ′′ t ′ )δ = −(α 1 · · · α m−1 )ρ H − (β m−1 · · · β 1 )ρ H .Choose ‘compensation’ relations (p, q) and (p ′ , q ′ ) in R + such that (p, q)δ =(α m−1 )ρ H +(β m−1 )ρ H and (p ′ , q ′ )δ = (α 1 · · · α m−2 )ρ H +(β −1m−2 · · · β 1)ρ H .The relations(sxα 1 · · · α m−2 wβ m−2 · · · β 1 y ′ tp, s ′ y ′′ t ′ q),(p ′ sxα m−1 wβ m−1 y ′ t, q ′ s ′ y ′′ t ′ ), and (pp ′ sxwy ′ t, qq ′ s ′ y ′′ t ′ ) ()are in ker ρ and precede (u, v) in the ≪-ordering. The following Malcevchain shows that the relation (u, v) is a Malcev consequence of the relations() and those in Q:sxα 1 · · · α m−1 wβ m−1 · · · β 1 y ′ t→ sxα 1 · · · α m−2 x L s L (p ′ ) L p ′ sxα m−1 wβ m−1 y ′ tt R (y ′ ) R β m−2 · · · β 1 y ′ t→ sxα 1 · · · α m−2 x L s L (p ′ ) L q ′ s ′ y ′′ t ′ t R (y ′ ) R β m−2 · · · β 1 y ′ t→ sxα 1 · · · α m−2 x L s L (p ′ ) L q L qq ′ s ′ y ′′ t ′ t R (y ′ ) R β m−2 · · · β 1 y ′ t→ sxα 1 · · · α m−2 x L s L (p ′ ) L q L pp ′ sxwy ′ tt R (y ′ ) R β m−2 · · · β 1 y ′ t→ sxα 1 · · · α m−2 x L s L (p ′ ) L q L pp ′ sxwβ m−2 · · · β 1 y ′ tqq R()by ()by ()→ sxα 1 · · · α m−2 x L s L (p ′ ) L q L qp ′ sxwβ m−2 · · · β 1 y ′ tpq R by Q #→ sxα 1 · · · α m−2 wβ m−2 · · · β 1 y ′ tpq R→ s ′ y ′′ t ′ qq R→ s ′ y ′′ t ′In either case, (u, v) ∈ ker ρ − S is a Malcev consequence of ≪-precedingelements of ker ρ plus relations from Q. Therefore, by induction on n(u, v), thesemigroup S has a Malcev presentation SgM⟨A | S ∪ Q⟩. 12Second stage. The reasoning thus far has reduced the ‘ordinary’ presentationSg⟨A | ker ρ⟩ for S to the Malcev presentation SgM⟨A | S ∪ Q⟩. However, the setof relations S ∪ Q is infinite. The next stage is to show that all relations in S ∪ Qare Malcev consequences of those in a still infinite — but simpler — set T.This section is rather technical, so a few motivational remarks will be madeimmediately after some definitions required later.by ()

Let K = (N ∪ {0}) 2n and N = Z n . Define δ ′ : K → K byn∑(a 1 , a 1 ′ , . . . , a n, a n) ′ [↦→ ai (u i , v i )δ + a i ′ (v i, u i )δ ] .i=1(Recall that R = {(u 1 , v 1 ), (v 1 , u 1 ), . . . , (u n , v n ), (v n , u n )}.)Define σ : K → N byLet δ ′′ : N → H be given by(a 1 , a ′ 1 , . . . , a n, a ′ n) ↦→ (a 1 − a ′ 1 , . . . , a n − a ′ n).(b 1 , . . . , b n ) ↦→n∑b i (u i , v i )δ.Recall that every relation in S is a concatenation of elements of R. It is obvious,therefore, that S ⊆ R # . However, although all elements of R lie in ker ρ F ,they may have non-zero image under δ. (Recall that (u, v)δ = uρ H −vρ H .) Suppose(u, v) = (c 1 , d 1 ) · · · (c k , d k ), with (c j , d j ) ∈ R and that this decompositioncontains a i instances of (u i , v i ) and ai ′ instances of (v i, u i ) for each i. Recordthis fact using a tuple T = (a 1 , a1 ′ , a 2, a2 ′ , . . . , a n, a n) ′ ∈ K. Notice that theimage of T under δ ′ coincides with the image of (u, v) under δ. So the tuple correspondingto any relation in ker ρ must also have image 0 H under δ ′ . Now, asthe contributions of each a i and ai ′ to the sum () are mutually inverse, one maypass to a tuple (a 1 −a1 ′ , . . . , a n−a n) ′ in N (using the mapping σ) and still be ableto obtain the image of T under δ ′ using the mapping δ ′′ . (Lemma formalizesthis notion.) The kernel of δ ′′ (in the group-theoretical sense) is a subgroup ofthe finitely generated abelian group N and is therefore itself finitely generated.The strategy is to pick a finite [semigroup] generating set for this kernel, pullthis set back to a set of tuples Y in K, and thus to find a particular set T consistingof relations formed by concatenating relations from R and trivial relations(a, a) such that the number of (u i , v i ) and (v i , u i ) is described by a tuple in Y.(Thus each relation in T has image 0 H under δ.) The aim is to express the tupleT as a positive (semigroup) sum of tuples ∑ j∈J y j with y j ∈ Y. The definitionof the relations in T then allows the construction of a Malcev chain from u tov in which there is a one-to-one correspondence between the steps of the chainand the y j in the sum of tuples.The details of the reasoning are, however, quite delicate: a number of technicaldifficulties arise in pulling back the generators of the kernel of δ ′′ to tuplesin K. Lemmata – show how to surmount these problems. . σδ ′′ = δ ′ .i=1Proof of 13. Let (a 1 , a ′ 1 , . . . , a n, a ′ n) ∈ K. Then(a 1 , a 1 ′ , . . . , a n, a n)σδ ′ ′′ = (a 1 − a 1 ′ , . . . , a n − a n)δ ′ ′′n∑= (a i − a i ′ )(u i, v i )δ==i=1n∑[a i (u i , v i )δ − a i ′ (u i, v i )δ]i=1n∑[a i (u i , v i )δ + a i ′ (v i, u i )δ]i=1= (a 1 , a ′ 1 , . . . , a n, a ′ n)δ ′ .

Therefore σδ ′′ = δ ′ . 13Now, Ker δ ′′ is a subgroup of N = Z n and so is finitely generated. Let X be afinite semigroup generating set for Ker δ ′′ . [This proof adheres to a notationaldistinction between ker ϕ, which denotes a congruence on the domain of ahomomorphism ϕ, and Ker ψ, which is a normal subgroup of the domain of agroup homomorphism ψ.]Let τ : N → K be defined by(b 1 , . . . , b n ) ↦→ (max{b 1 , 0}, max{−b 1 , 0}, . . . , max{b n , 0}, max{−b n , 0}).Observe that τσ = id N . This observation and Lemma will together showthat τ is ‘almost’ an inverse of σ. Use τ to pull back X into K as follows: LetY ′ = Xτ. Let Y ′′ ⊆ K be the set{(1, 1, 0, 0, . . . , 0, 0), (0, 0, 1, 1, . . . , 0, 0), . . . , (0, 0, 0, 0, . . . , 1, 1)},and let Y = Y ′ ∪ Y ′′ .Unfortunately, the composition στ is not the identity mapping. However, itis close enough for the purposes of this proof, in a sense made precise by thefollowing lemma: . For (a 1 , a ′ 1 , . . . , a n, a ′ n) ∈ K,(a 1 , a ′ 1 , . . . , a n, a ′ n) − (a 1 , a ′ 1 , . . . , a n, a ′ n)στ ∈ Mon ⟨ Y ′′⟩ .Proof of 14. Let (. . . , a i , ai ′ , . . .) ∈ K. Then(. . . , a i , a ′ i , . . .)στ = (. . . , a i − a ′ i , . . .)τ= (. . . , max{a i − a ′ i , 0}, max{a′ i − a i, 0}, . . .).If a i ⩾ ai ′, then this gives (. . . , a i − ai ′ , 0, . . .), and(. . . , a i , a i ′ , . . .) − (. . . , a i − a i ′ , 0, . . .) = a′ i (. . . , 1, 1, . . .) + . . . .If a i ⩽ a ′ i , then this gives (. . . , 0, a′ i − a i, . . .), and(. . . , a i , a ′ i , . . .) − (. . . , 0, a′ i − a i, . . .) = a i (. . . , 1, 1, . . .) + . . . .Reasoning thus for each i gives the result. 14Similarly, τ is ‘close’ to being a homomorphism: . For (b 1 , . . . , b n ) and (c 1 , . . . , c n ) in N,(b 1 , . . . , b n )τ + (c 1 , . . . , c n )τ − (b 1 + c 1 , . . . , b n + c n )τ ∈ Mon ⟨ Y ′′⟩ .Proof of 15. Let (. . . , b i , . . .) and (. . . , c i , . . .) be members of N. Then(. . . , b i , . . .)τ + (. . . , c i , . . .)τ= (. . . , max{b i , 0}, max{−b i , 0}, . . .) + (. . . , max{c i , 0}, max{−c i , 0}, . . .)= (. . . , max{b i , 0} + max{c i , 0}, max{−b i , 0} + max{−c i , 0}, . . .).Consider the following four cases:

. b i , c i ⩾ 0. This gives (. . . , b i , . . .)τ + (. . . , c i , . . .)τ = (. . . , b i + c i , 0, . . .), and(. . . , b i + c i , 0, . . .) − (. . . , b i + c i , . . .)τ= (. . . , b i + c i , 0, . . .) − (. . . , b i + c i , 0, . . .)= (. . . , 0, 0, . . .).. b i , c i ⩽ 0. This gives (. . . , b i , . . .)τ + (. . . , c i , . . .)τ = (. . . , 0, −b i − c i , . . .),and(. . . , 0, −b i − c i , 0, . . .) − (. . . , b i + c i , . . .)τ= (. . . , 0, −b i − c i , 0, . . .) − (. . . , 0, −b i − c i , . . .)= (. . . , 0, 0, . . .).. b i ⩾ 0, c i ⩽ 0. Now split into two sub-cases:(a) |b i | ⩾ |c i |. This gives (. . . , b i , . . .)τ + (. . . , c i , . . .)τ = (. . . , b i , −c i , . . .),and(. . . , b i , −c i , 0, . . .) − (. . . , b i + c i , . . .)τ= (. . . , b i , −c i , 0, . . .) − (. . . , b i + c i , 0, . . .)= (. . . , b i − b i − c i , −c i . . .)= (. . . , −c i , −c i , . . .).(Observe that −c i ⩾ 0.)(b) |b i | ⩽ |c i |. This gives (. . . , b i , . . .)τ + (. . . , c i , . . .)τ = (. . . , b i , −c i , . . .),and(. . . , b i , −c i , 0, . . .) − (. . . , b i + c i , . . .)τ= (. . . , b i , −c i , 0, . . .) − (. . . , 0, −(b i + c i ), . . .)= (. . . , b i , −c i + b i + c i . . .)= (. . . , b i , b i , . . .).. b i ⩽ 0, c i ⩾ 0. This is symmetric to case iii.Apply the four cases above to each i to complete the proof. 15Finally, although Ker δ ′ is not guaranteed to be a subset of Mon⟨Y⟩, anyelement of Ker δ ′ differs from some element of Mon⟨Y⟩ only by an element ofMon⟨Y ′′ ⟩: . If (a 1 , a ′ 1 , . . . , a n, a ′ n)δ ′ = 0 H , then there exists y ∈ Mon⟨Y ′′ ⟩ suchthat (a 1 , a ′ 1 , . . . , a n, a ′ n) + y ∈ Mon⟨Y⟩.Proof of 16. Let (a 1 , a ′ 1 , . . . , a n, a ′ n)δ ′ = 0 H . Lemma shows that the image of(a 1 , a ′ 1 , . . . , a n, a ′ n) under σ lies in Ker δ ′′ . So(a 1 , a ′ 1 , . . . , a n, a ′ n)σ = ∑ j∈Jx j for some x j ∈ X,((a 1 , a 1 ′ , . . . , a ∑n, a n)στ ′ = x j)τ,j∈J(a 1 , a ′ 1 , . . . , a n, a ′ n)στ + y = ∑ j∈J(x j τ) where y ∈ Mon ⟨ Y ′′⟩ , by Lemma ,(a 1 , a ′ 1 , . . . , a n, a ′ n) + y = ∑ j∈J(x j τ) + z where z ∈ Mon ⟨ Y ′′⟩ , by Lemma ,∈ Mon⟨Y⟩ ,

and this completes the proof. 16Let T consist of all relations in (R ∪ {(a, a) : a ∈ A}) + , containing, for some(a 1 , a1 ′ , . . . , a n, a n) ′ ∈ Y, exactly a i instances of (u i , v i ) and ai ′ instances of(v i , u i ), for each i. Observe that T contains Q since Y contains Y ′′ . Notice thatthe definition of T makes no mention of canonical decompositions. When a relationin T used to rewrite a word over A, these a i instance of u i change to v i ,and the ai ′ instances of v i change to u i , and no other letters alter. Call theseunchanged intermediate letters ‘padding’.Suppose (u, v) ∈ T. Then obviously (u, v) ∈ ker ρ F . Also, (u, v) ∈ ker ρ H ,since(u, v)δ = (a 1 , a 1 ′ , . . . , a n, a n)δ ′ ′ = 0 H ,where (a 1 , a1 ′ , . . . , a n, a n) ′ ∈ Y. So T ⊆ ker ρ.Define χ : S → K as follows: w ∈ S maps to (a 1 , a1 ′ , . . . , a n, a n), ′ wherea i is the number of (u i , v i ) in the canonical decomposition of w, and ai ′ is thenumber of (v i , u i ). . Every relation in S is a Malcev consequence of those in T.Proof of 17. Let (u, v) ∈ S. Then (u, v) ∈ ker ρ F ∩ker ρ H , so (u, v)χ ∈ Ker δ ′ . Lety ∈ Mon⟨Y ′′ ⟩ be such that (u, v)χ+y ∈ Mon⟨Y⟩. Suppose y = (b 1 , b 1 , . . . , b n , b n ).Let(s, t) = (u 1 , v 1 ) b 1(v 1 , u 1 ) b1 · · · (u n , v n ) b n(v n , u n ) b n. ()Observe that (s, t) ∈ Q # .Suppose that (u, v)χ + y = ∑ pj=1 y j, where y j ∈ Y. Construct a Malcev T-chain from u to v as follows. First of all, insert t and transform it to s usingrelations from Q ⊆ T:u → utt R → ust R .()The component a i of the tuple (u, v)χ + y describes the number of (u i , v i )that appear in the canonical decomposition of (u, v) concatenated with the decomposition() of (s, t). A similar statement applies to ai ′ and (v i, u i ). Putanother way, the components a i and ai ′ of the tuple (u, v)χ + y describes thenumber of subwords u i and v i of us that must be changed to v i and u i , respectively,in order to transform us to vt. As (u, v)χ + y = ∑ pj=1 y j, the sum of thea i and ai ′ components of the y j also gives the number subwords of each typethat must be changed.Construct a Malcev chain from us to vt by defining the jth step in the chain(where j = 1, . . . , p) as follows. Suppose y j = (a 1 , a1 ′ , . . . , a n, a n) ′ and that thefirst j − 1 steps have transformed us to w. For each i, find a i subwords u i andai ′ subwords v i of w that have not been changed in the chain thus far. The jthstep consists of changing those words u i to v i and v i to u i . By the definition ofT, a relation exists that permits this step. Furthermore, by the comments in thelast paragraph, the word left after the pth step is vt.Concatenate the Malcev chain () with the one just constructed and appendvtt R → vto obtain a Malcev T-chain from u to v. 17Third stage. The proof thus far has shown that S has a Malcev presentationSgM⟨A | T⟩. The set T is still infinite. This third and final stage shows that afinite subset U of T will suffice in a Malcev presentation for S.

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:

. Is every direct product of a free group and a nilpotent groupMalcev coherent?The class of coherent groups is closed under forming finite extensions. Thereforethe direct product of a free group and a virtually polycyclic group is alsocoherent. This immediately provokes the following question: . Is every direct product of a free group and a virtually abeliangroup Malcev coherent?It is not yet known whether the class of Malcev coherent groups is closed underforming finite extensions. [The author opines that this is the most importantunsolved problem in the theory of Malcev presentations.] Thus it is conceiveablethat these two questions may have different answers if one replaces ‘free’by ‘virtually free’.For a general survey of the current state of knowledge of Malcev presentations,see [Cai].The author gratefully acknowledges the support of the CarnegieTrust for the Universities of Scotland. The author thanks the anonymous refereefor many constructive comments, especially regarding the exposition ofthe proof of Propostion . [Bau] G. Baumslag. ‘A remark on generalized free products’. Proc. Amer. Math.Soc., (), pp. –. doi: ./.[BMa] L. G. Budkina & A. A. Markov. ‘F-semigroups with three generators’.Mat. Zametki, (), pp. –. [In Russian. See [BMb] for a translation.].[BMb] L. G. Budkina & A. A. Markov. ‘F-semigroups with three generators’.Math. Notes, (), pp. –. [Translated from the Russian.].[Cai] A. J. Cain. ‘Automatic structures for subsemigroups of Baumslag–Solitarsemigroups’.[Cai] A. J. Cain. Presentations for Subsemigroups of Groups. Ph.D. Thesis, Universityof St Andrews, . Available from: www-groups.mcs.st-and.ac.uk/~alanc/publications/c_phdthesis/c_phdthesis.pdf.[Cai] A. J. Cain. ‘Malcev presentations for subsemigroups of groups — a survey’.In C. M. Campbell, M. Quick, E. F. Robertson, & G. C. Smith, eds,Groups St Andrews 2005 (Vol. ), no. in London Mathematical SocietyLecture Note Series, pp. –, Cambridge, . Cambridge UniversityPress.[CP] A. H. Clifford & G. B. Preston. The Algebraic Theory of Semigroups (Vol. II).No. in Mathematical Surveys. Amer. Math. Soc., Providence, R.I., .[CRRa] A. J. Cain, E. F. Robertson, & N. Ruškuc. ‘Subsemigroups of groups: presentations,Malcev presentations, and automatic structures’. J. Group Theory,, no. (), pp. –. doi: ./jgt...[CRRb] A. J. Cain, E. F. Robertson, & N. Ruškuc. ‘Subsemigroups of virtuallyfree groups: finite Malcev presentations and testing for freeness’.

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