Let $*$ be a binary operation on a set $S$. We define a binary relation $R$ on $S$ by: $xRy$ iff $\exists z: x*z=y$. If $*$ is associative, then the relation $R$ is transitive. My question is whether the converse is true.
Asked
Active
Viewed 413 times
3
-
Assuming * is close and identity and inverses exist, then let $(ax_1)x_2=c$, $ax_1=b$ then $aRb, bRc \rightarrow aRc$ means there is $x_3$ such that $ax_3=c$. So then $(ax_1)x_2=ax_3$ implies $x_1x_2=x_3$ because a is invertible and you multiply inverse of a on both side of equation to cancel it out. Then $(ax_1)x_2=a(x_1x_2)$. – user614287 Apr 19 '19 at 20:08
1 Answers
3
No. For instance, let $S=\{0,1\}$ and let $*$ be the operation $$0*0=1,$$ $$0*1=0,$$ $$1*0=1,$$ $$1*1=1.$$ Then the relation $R$ is transitive (it is just the usual order relation $\leq$), but $*$ is not associative since $$(0*0)*0=1\neq 0=0*(0*0).$$
Eric Wofsey
- 330,363