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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?

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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 $e$.

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  • $\begingroup$ 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

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