I am struggling with the following theorem about semigroups, so I was hoping someone could give me a hand.

The theorem states: "Let $S$ be an arbitrary semigroup such that for every $a\in{}S$ it holds $a\in aS^2$ and for every $a,x\in S$ the condition $a^2x=a$ implies that $x^2a=x$. Then for every $a,b\in S$ we have that $a\in abSa$"

I will show you my attempt. So far, I was only able to prvef that $S$ is completely regular. Indeed, consider any element $a\in S$. Then $a=a^2x$ for some $x\in S$, which implies that $x=x^2a$. But then \begin{align*} a&=a^2x=a^2x^2a=(a^2x)xa=axa,\\ x&=x^2a=x^2a^2x=(x^2a)ax=xax \end{align*} So element $a$ is regular (and therefore $S$ is a regular group) and if $a=a^2x$, then $x$ is the inverse of $a$. Now consider the element $a^2$. From the hypothesis, we know, that $a^2=a^4y$ which implies $y=y^2a^2$. Also from the previous, we know, that $y$ is inverse of $a^2$, so we get $$a=a^2x=(a^2ya^2)x=a^2y(a^2x)=a^2ya=a^2(y^2a^2)a=(a^2y^2a)a^2$$ This means that $a\in Sa^2$. So, now we have $a\in a^2S\land a\in Sa^2$ which implies, that $a$ is a completely regular element, therefore $S$ is a completely regular semigroup.

I don't know how to continue from here, so I would appreciate any help. Thanks

  • $\begingroup$ My idea was to prove, that $a\mathscr{R}ab$ for every $a,b\in{}S$. Then if $x$ denotes the inverse of $a$ that commutes with $a$, we have $a=a^2x=abcax=abcxa=ab(cx)a$, for some $c\in{}S$. But I was not able to prove that $a\mathscr{R}ab$. $\endgroup$ Sep 17, 2013 at 0:21
  • $\begingroup$ FYI, all those braces to separate things were unnecessary: a \in n produces $a\in b$ and a \land b produces $a\land b$. $\endgroup$
    – dfeuer
    Sep 17, 2013 at 0:26
  • $\begingroup$ You should mention that this is a follow-up to your question math.stackexchange.com/questions/486575/… and that you are still trying to do this exercise. However, since there are 5 equivalent conditions to prove, you should tell which implications between the 5 conditions you already proved. $\endgroup$
    – J.-E. Pin
    Sep 17, 2013 at 3:05
  • $\begingroup$ @J.-E.Pin That's true. So far, I got $\alpha\Rightarrow\beta$, $\beta\Rightarrow\gamma$, $\alpha\Rightarrow\delta$, $\delta\Rightarrow\alpha$ $\endgroup$ Sep 17, 2013 at 7:27


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