I have a semigroup $G$ (a set with associative binary operation) such that for all $a,b\in G$ there exists $x,y\in G$ such that $ax=ya=b$. Is this property enough to show that $G$ is a group, and if so, how?
Semigroup with "transitive" operation is a group?
6
$\begingroup$
group-theory
semigroups
1 Answers
9
Yes. This is a standard exercise.
First, let $a\in G$ be arbitrary. Then there exists $e\in G$ such that $ae=a$ (using the first condition with $a=b$). Now let $b\in G$. Then there exists $z\in G$ such that $za=b$. Therefore, $be=(za)e=z(ae) = za = b$. Thus, $e$ is a right identity for $G$.
From the fact that $ax=e$ is always solvable it follows that every element has a right inverse. Since your semigroup has a right identity and right inverses, then it is a group.