Malcev presentations for subsemigroups of direct products of ...

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.()