On a combinatorial property of Fibonacci semigroup
Let \(A^ +\) be a free semigroup over a two-letter alphabet \(A = \{a,b\}\), \(\varphi : A^ + \to A^ +\) be the morphism defined by \(\varphi(a) = ab\), \(\varphi(b) = a\). Let \(f_ 1 = a\) and \(f_{n + 1} = \varphi(f_ n)\). Then \(| f_ 1|, | f_ 2 |, \dots, | f_ n|, \dots\) is the sequence of Fibonacci numbers. By iterating the morphism \(\varphi\), starting with \(a\), we can define a (unique) infinite Fibonacci word \(f = abaababaabaab \dots\). Denote by \({\mathcal F} (f)\) the set of all factors of \(f\). The Fibonacci semigroup \(\mathcal F\) is, by definition, the Rees quotient of \(A^ +\) by the ideal \(J(f) = A^ + - {\mathcal F}(f)\) of nonfactors of \(f\). We say that a semigroup \(S\) has property \(P^*_ n\), \(n \geq 2\), if, given elements \(x_ 1, \dots, x_ n\) of \(S\), at least two of the \(n!\) products of these elements coincide. \textit{A. Restivo} proved that the Fibonacci semigroup \(\mathcal F\) has property \(P^*_ 8\) [see Permutation properties and the Fibonacci semigroup, Semigroup Forum (to appear)]. The author proves that \(\mathcal F\) has property \(P^*_ 4\) but does not have property \(P^*_ 3\). In fact, the six possible products of three elements \(a, b, abaaba\) of \({\mathcal F}(f)\) belong to \({\mathcal F}(f)\) and are different. This is the best possible result.
scientific article; zbMATH DE number 500643