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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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1 Answers1

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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 $e$.

kobe
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  • I am having trouble seeing how "some power of $x$ equals $e$" – Learnmore Jan 15 '17 at 04:14
  • @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. – kobe Jan 15 '17 at 04:29
  • Right !I missed that;Thank you – Learnmore Jan 15 '17 at 04:42
  • If $x^n$ is idempotent, then $x^n = e$ since the only idempotent in $G$ is $e$, right? Does it implies that $G$ is a group? – lap lapan Jan 18 '21 at 22:01
  • @kobe Let $n=3$. Then, $x^3=e$. Now, $x^{-1}=x^2$. Thus, the inverse of any $x$ in $G$ is $x^2$. Like this? – lap lapan Jan 18 '21 at 22:11
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    @user795084 Yes, if $x^n$ is idempotent, then since $e$ is idempotent, uniqueness gives $x^n = e$. Then $x$ has inverse $x^{n-1}$. Note, however, that $n$ depends on $x$. So your example only makes sense for some $x$, not for all $x$. – kobe Jan 18 '21 at 22:29
  • @kobe Nice, I got it now. Thanks Sir! – lap lapan Jan 18 '21 at 22:33
  • @kobe Now, does it implies that every $x$ in $G$ has an inverse? Since you say that $n$ depends on $x$. – lap lapan Jan 18 '21 at 23:28
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    @user795084 given $x\in G$, there exists an $n$ such that $x^n =e $. Thus $x$ is invertible with inverse $x^{n-1}$. Since $x$ is arbitrary, every element of $G$ has an inverse. – kobe Jan 18 '21 at 23:41