Malcev presentations for subsemigroups of direct products of ...

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 ′′