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Theorem: Let $S$ be an inverse semigroup, and let $x,y\in S$ and $e,f\in E_{S}$ then

  1. $x\mathcal{L}y$ if and only if $x^{-1}x=y^{-1}y$

  2. $x\mathcal{R}y$ if and only if $xx^{-1}=yy^{-1},$

where $E_S$ denotes the set of idempotents of $S$, and $\mathcal{L}$, $\mathcal{R}$ are Green's relations.

For the forward of 1) I argue that if $x\mathcal{L}y$ then $L_{x}=L_{y}$. And since S is an inverse semigroup, S is regular and hence each $\mathcal {L}$-class contains a unique idempotent.

$x^{-1}x$ is idempotent in $L_{x}$ (why?), and $y^{-1}y$ is also idempotent in $L_{y}$ (why?). Hence $x^{-1}x=y^{-1}y$ Since each class contains a unique idempotent.

To be sincere, I did not convince myself about my proof, but I know somebody will surely point out my mistakes and corrected me.


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up vote 4 down vote accepted

Your proof is flawed, but not very much.

You say, "And since S is an inverse semigroup, S is regular and hence each $\mathscr{L}$-class contains a unique idempotent."

No, it's not true that in a regular semigroup each $\mathscr{L}$-class contains a unique idempotent. The theorem is this:

Theorem 1. A semigroup $S$ is regular if and only if each $\mathscr L$- and $\mathscr R$-class contains an idempotent.

There's nothing about uniqueness here, and indeed those idempotents may not be unique. (Try finding an example!) However, there is another theorem:

Theorem 2. A semigroup $S$ is inverse if and only if each $\mathscr L$- and $\mathscr R$-class contains a unique idempotent.

So the uniqueness you're using follows from inverseness of $S$, not from regularity.

Now, you're asking why $x^{-1}x$ is an idempotent in $L_x$. But this is an easy exercise. We have $$(x^{-1}x)^2=(x^{-1}x)(x^{-1}x)=(x^{-1}xx^{-1})x=x^{-1}x.$$

This only uses the definition of the inverse. Now why is it in $L_x$? It is enough to show that there exist $a,b\in S^1$ such that $x=ax^{-1}x$ and $x^{-1}x=bx.$ Take $a=x$ and $b=x^{-1}.$

Now the other direction. Since $x^{-1}x$ and $y^{-1}y$ are equal and in the same $\mathscr L$-class as $x$ and $y,$ we have $$S^1x=S^1x^{-1}x=S^1y^{-1}y=S^1y$$ and so $x\mathscr L y.$


I will give the proofs of theorems 1 and 2 here. First we will note the following fact.

Fact. Let $S$ be a regular semigroup and $x\in S.$ Then $S^1x=Sx$ and $xS^1=xS.$

I will leave the simple proof to you. This fact has the following corollary.

Corollary. Let $S$ be a regular semigroup and $x,y\in S.$ Then $$(1)\;\;x\mathscr L y \iff Sx=Sy$$ and $$(2)\;\;\;\;x\mathscr R y \iff xS=yS.$$

Proof of theorem 1. Suppose $S$ is a regular semigroup. Let $x\in S.$ Then there is $y$ such that $xyx=x$ and we have $(yx)^2=yx.$ We note that $x=x(yx)$ implies $Sx\subseteq S(yx)$ and $(yx)=y(x)$ implies $S(xy)\subseteq Sx.$ Therefore $$Sx=S(yx)$$ and $$x\mathscr L (yx).$$ Hence $L_x$ contains an idempotent, namely $yx$. We prove that the idempotent $xy$ is contained in $R_x$ similarly.

Conversely, suppose each $\mathscr L$- and $\mathscr R$-class contains an idempotent. Let $x\in S.$ We have idempotents $e^2=e\in L_x$ and $f^2=f\in R_x.$ There exist $a,b,c,d\in S^1$ such that $$x=ae,\;\;\;e=bx,\;\;\;x=fc,\;\;\;f=xd.$$ (Notice that we can't write $a,b,c,d\in S.$ We don't know $S$ is regular yet.)

Now we have $$x=ae=aee=xe=xbx=fcbx=ffcbx=fxbx=x(dxb)x,$$ where $b,d\in S^1.$ If $b,d\in S,$ then $dxb\in S$ and we are done. If $b\not\in S,$ then $b=1.$ Then if $d\in S,$ we have $dxb=dx\in S$ and we are done. But if also $d=1,$ then $dxb=x\in S$ and we are done. Similarly, nothing bad happens if $d=1.$ $\square$

Remark 1. Note that we could have been satisfied with $x=xbx.$ Indeed, then if $b\in S,$ the proof is done. If $b=1,$ then $x=xx=xxx$ and we are also done. I posted the previous calculation because the formula we obtain doesn't prefer either of the elements $b$ and $d.$

Remark 2. Note also that we have $$(bxd)x(bxd)=b(x(dxb)x)d=bxd.$$

Proof of theorem 2. Suppose $S$ is an inverse semigroup. It is therefore a regular semigroup and from theorem 1, we know that for every $x\in S$ we have idempotents $e^2=e\in L_x$ and $g^2=g\in R_x.$ We need to prove that they're unique.

Suppose $f^2=f\in S$ and $e\mathscr L f.$ Then there exist $a,b\in S$ such that $$ae=f,\;\;\;bf=e.$$ But then $$efe=bffe=bfe=ee=e$$ and analogously $$fef=f.$$ Therefore $f$ is an inverse of $e.$ But also $e$ is an inverse of $e$ because $eee=e.$ Since in an inverse semigroup we have unique inverses, this gives us that $e=f.$ Analogously, we prove that idempotents in $\mathscr R$-classes are unique.

Conversely, suppose there are unique idempotents in $\mathscr L$- and $\mathscr R$-classes of $S.$ From theorem 1, we know that this implies the regularity of $S.$ This ensures the existence of inverses. We need to prove their uniqueness. Let $x\in S$ and let $x_1,x_2\in S$ be inverses of $x.$ It should be clear by now that it implies $$(x_1x)\mathscr L (x_2x),\;\;\;(xx_1)\mathscr R (xx_2).$$ But those elements are idempotents so we must have $$x_1x=x_2x,\;\;\;xx_1=xx_2.$$ Therefore $$x_1=x_1xx_1=x_2xx_1=x_2xx_2=x_2.$$ We proved that inverses are unique in $S,$ which ends the proof of the theorem. $\square$

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Thank you for you answer. I do not know about the theorem S is an inverse semigroup iff each $\mathcal{L}$ and $\mathcal{R}$-classes contains unique idempotent. If you can help with the proof, I will be glad. – Hassan Muhammad Mar 25 '12 at 17:22
@HassanMuhammad Do you have any textbook available? If you want to learn about inverse semigroups, you just need one of Howie's books. Either An introduction to semigroup theory or Fundamentals of semigroup theory. Can you find them? It would be more beneficial for you to have a look at the proofs there, as you will see them in context. – user23211 Mar 25 '12 at 17:26
@HassanMuhammad If you can't access any books, please tell me and I will prove those proposition for you. – user23211 Mar 25 '12 at 17:39
I can't access any book on semigroup now. I will be very happy to see the proof, because my mates will never accept my proof without the proof of that proposition in the tutorial classes. Thanks – Hassan Muhammad Mar 25 '12 at 17:44
I have only some old lecture notes without the proof of the propositions. – Hassan Muhammad Mar 25 '12 at 17:59

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