1976•Proceedings of the American Mathematical SocietyRequires access

Conditions for the Commutativity of Semigroups

G. Kowol

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Abstract

Let $S$ be a semigroup. Then by a theorem of Tully [7]: $S$ is a commutative semigroup iff $ab = {b^n}{a^m}$ for all $a,b \in S$ ($m,n \geqslant 1$, fixed). We prove the following: $S$ is a commutative semigroup iff $ab = {b^{n(a,b)}}{a^{m(a,b)}}$ for all $a,b \in S$, where one of the exponents $n(a,b)$ and $m(a,b)$ is constant and the other is independent of $a$ or $b$.

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What this paper is about

Let $S$ be a semigroup. Then by a theorem of Tully [7]: $S$ is a commutative semigroup iff $ab = {b^n}{a^m}$ for all $a,b \in S$ ($m,n \geqslant 1$, fixed). We prove the following: $S$ is a commutative semigroup iff $ab = {b^{n(a,b)}}{a^{m(a,b)}}$ for all $a,b \in S$, where one of the exponents $n(a,b)$ and $m(a,b)$ is constant and the other is independent of $a$ or $b$.

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Available abstract

Let $S$ be a semigroup. Then by a theorem of Tully [7]: $S$ is a commutative semigroup iff $ab = {b^n}{a^m}$ for all $a,b \in S$ ($m,n \geqslant 1$, fixed). We prove the following: $S$ is a commutative semigroup iff $ab = {b^{n(a,b)}}{a^{m(a,b)}}$ for all $a,b \in S$, where one of the exponents $n(a,b)$ and $m(a,b)$ is constant and the other is independent of $a$ or $b$.

Key concepts: Semigroup, Commutative property, Mathematics, Constant (computer programming), Pure mathematics, Discrete mathematics, Combinatorics, Computer science

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