Let $G$ be a finite semi-group with identity such that it has only one idempotent.Is $G$ a group?

It only remains to show that for any $a\in G$ $\exists b\in G$ such that $ab=ba=e$ where $e$ is the identity of $G$

Also $e$ is the only idempotent of $G$ .How to proceed next?


Show that for every $x\in G$, there is an $n\in \Bbb N$ such that $x^n$ is idempotent. Then you can claim that for every $x\in G$, some power of $x$ equals $e$.

  • $\begingroup$ I am having trouble seeing how "some power of $x$ equals $e$" $\endgroup$ – Learnmore Jan 15 '17 at 4:14
  • $\begingroup$ @learnmore it's given that $G$ has only one idempotent, which we know to be $e$. So if $x^n$ is idempotent, then it must equal $e$ by uniqueness. $\endgroup$ – kobe Jan 15 '17 at 4:29
  • $\begingroup$ Right !I missed that;Thank you $\endgroup$ – Learnmore Jan 15 '17 at 4:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.