Sum-free subsets of right cancellative semigroups (Q5949036)

For a natural number \(k\geq 2\) let \(\rho=\rho(k)\) be the smallest natural number which does not divide \(k-1\). A subset \(A\) of a right cancellative semigroup \((S,\cdot)\) is \(k\)-product-free if \(A\cap A^k=\emptyset\) (if the operation on \(S\) is denoted by \(+\) we say that it is \(k\)-sum-free).NEWLINENEWLINENEWLINEIn this paper it is proved that if \(A\) is \(k\)-product-free then there exists \(s\in S\) such that the sets \(A,As,As^2,\dots,A^{\rho-1}\) are pairwise disjoint. This result is connected with a conjecture of \textit{N. J. Calkin} and \textit{P. Erdős} [see Math. Proc. Camb. Philos. Soc. 120, No. 1, 1-5 (1996; Zbl 0866.11019)] on \(k\)-sum-free subsets of the semigroup \((\mathbb{N},+)\). The authors of this paper gave an affirmative answer to this conjecture [see \textit{T. Łuczak} and \textit{T. Schoen}, J. Number Theory 66, No. 2, 211-224 (1997; Zbl 0884.11018)]. They proved that if \(A\) is a \(k\)-sum free subset of \((\mathbb{N},+)\) then \(\overline d(A)\leq 1/\rho(k)\), where \(\overline d(A)=\limsup_{n\to\infty}\overline d_n(A)\), with \(\overline d_n(A)=|A\cap\{1,\dots,n\}|/n\).