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I have been looking at this for hours now. Why is it true that idempotents of an inverse semigroup commute? It seems like this should be straightforward but I just can't get it.

Any help is greatly appreciated.

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    $\begingroup$ In future please specify which definitions you're using, since, for instance, some authors define an inverse semigroup as a regular semgroup whose idempotents commute. $\endgroup$
    – Shaun
    Jan 7, 2015 at 8:50
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    $\begingroup$ you could just give the OP the benefit of the doubt that s/he is not aware of this fact, assume s/he thinks that the definitions are universal (as most definitions are or almost are), and that this was not a lazy act on part of the person asking the question. You could have just as easily asked for clarity. $\endgroup$
    – Squirtle
    Jan 23, 2018 at 23:08

1 Answer 1

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Assume that $S$ is an inverse semigroup, and let $e, f\in S$ be arbitrary two idempotents. Then $$(ef)(f(ef)^{-1}e)(ef)=ef^2(ef)^{-1}e^2f=ef(ef)^{-1}ef=ef$$ $$(f(ef)^{-1}e)(ef)(f(ef)^{-1}e)=f(ef)^{-1}e^2f^2(ef)^{-1}e=f((ef)^{-1}ef(ef)^{-1})e=f(ef)^{-1}e.$$ Therefore, by the uniqueness of inverses, $f(ef)^{-1}e=(ef)^{-1}$. It follows that $$ (ef)^{-2}=(f(ef)^{-1}e)^2=f((ef)^{-1}ef(ef)^{-1})e=f(ef)^{-1}e=(ef)^{-1}, $$ i.e. $(ef)^{-1}$ is an idempotent and so $(ef)^{-1}=ef$. By symmetry, $fe$ is also an idempotent.

So, $$ (ef)(fe)(ef)=efef=ef,\ (fe)(ef)(fe)=fefe=fe, $$ and so $fe=(ef)^{-1}=ef$, as required.

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  • $\begingroup$ Nice, can you recommend a text on semi-groups that contains this stuff? $\endgroup$
    – Asinomás
    Jan 6, 2015 at 20:43
  • $\begingroup$ I'd recommend John Howie's book called "Fundamentals of semigroup theory" from 1995. $\endgroup$ Jan 6, 2015 at 20:45
  • $\begingroup$ That book looks nice, could you recommend another one plase? One on the cheaper side of things? $\endgroup$
    – Asinomás
    Jan 6, 2015 at 20:50
  • $\begingroup$ Lawson's Inverse Semigroups: The Theory of Partial Symmetries fits the bill, @JorgeFernández, but beware: Lawson's use of $\mathbf{d}$ and $\mathbf{r}$ was switched - I think to ease notation at the time. $\endgroup$
    – Shaun
    Jan 7, 2015 at 8:40

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