Consider any completely regular semigroup $S$. I would like to prove, that any $a,b\in{S}$ satisfies the identities $ab=a(ba)^0b=a(b^0a^0)^0b$, $a,b\in{}S$.

So far, I was able to prove only the first equation and I would appreciate any help with the second one. I will show you how I proved the first one:

First, consider the "inverse product formula" for completely regular semigroups, which states $$(ab)^{-1}=(ab)^0b^{-1}(ba)^0a^{-1}(ab)^0$$ By using this, we can get that \begin{align}ab&=ab(ab)^0=ab(ab)^{-1}ab=ab(ab)^0b^{-1}(ba)^0a^{-1}(ab)^0ab=\\&= abb^{-1}(ba)^0a^{-1}ab=ab^0(ba)^0a^0b=a(ba)^0b \end{align}

which proves the first identity. (In the last step, I used the rule $a^0(ab)^k=(ab)^k=(ab)^kb^0,\ k\in\mathbb{Z}$.) But I am struggeling with the second one, so I would appreciate any help :-)

  • $\begingroup$ What is the meaning of $a^0$ ? $\endgroup$
    – J.-E. Pin
    Sep 8, 2013 at 22:37
  • $\begingroup$ @J.-E.Pin: $a^0$ is the identity of $H_a$ where $H_a$ is the class of Rees $\mathcal{H}$ equivalence containing $a$. $\endgroup$ Sep 11, 2013 at 23:35

1 Answer 1


I figured it out, we can use the same "strategy" as with the first identity, the key is just being able to cleverly rewrite the product $ab$. Here is how $$ab=a(a^0b^0)b=a(a^0b^0)^0b=a(a^0b^0)(a^0b^0)^{-1}(a^0b^0)b$$

Now we can again use the "inverse product formula" for semigroups for $(a^0b^0)^{-1}$ and some similar steps as in the proof of the first identity. Eventually, we get that $$ab=a(b^0a^0)^0b$$


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