Malcev presentations for subsemigroups of direct products of ...

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