# Do the idempotents in an inverse semigroup commute?

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.

• 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. Jan 7, 2015 at 8:50
• 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. Jan 23, 2018 at 23:08

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.
• 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. Jan 7, 2015 at 8:40