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Let $a,b$ be elements of a regular semigroup $S$. Then $(a,b) \in \mathcal L$ iff there exist an inverse $a'$ of $a$ and $b'$ of $b$ such that $a' a = b'b$.

I have done the converse parts: if there exist an inverse $a'$ of $a$ and $b'$ of $b$ such that $a' a = b'b$ , then $(a,b) \in \mathcal L$.

Suppose $(a,b) \in \mathcal L$, so $a,b$ belongs the same $\mathcal D-$ class says $D$ and $a,b$ are regular, so every $\mathcal R$ class in $D$ contains an idempotent , it follows that $\mathcal R_b$ contains an idempotent says $e$. how to proceed further .

Any help would be appreciated, Thank you

user120386
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1 Answers1

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Suppose that $(a,b) \in \mathcal{L}$. Then there exists an invese $a'$ of $a$ such that $a'a$ is an idempotent in $L_a = L_b$. You know that every $\mathcal{R}$ contains an idempotent says $e$, which implies that $\exists \ b' \in R_{a'a} \cap L_e$ such that $b'b=a'a$. The reason of $\exists \ b' \in R_{a'a} \cap L_e$ is explained as follows.

Suppose that $e \in R_b$. We get $(e, b) \in \mathcal{R}$, and $eb=b$, $bx=e$ for $x \in S^1$, where $S^1$ is obtained by adding $1$ to $S$ if necessary. Moreover, $(a'a,b) \in \mathcal{L}$ implies that $ba'a = b$, $yb = a'a$ for $y \in S^1$.

Take $b' = a'axe$. You could get $bb'b=b$, $b'bb'=b'$, $bb'=e$, $b'b=a'a$. So $b' \in R_{a'a} \cap L_e$.

bing
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  • How $\exists b' \in \mathcal R_{a'a} \cap \mathcal L_e $ such that $b'b = a'a$ – user120386 Nov 02 '16 at 08:49
  • Thanks for the prompt reply – user120386 Nov 02 '16 at 08:50
  • @ user120386 I have added the reason. – bing Nov 02 '16 at 09:56
  • At last , I have doubt on $b'b= a'a$. Now $b'b = a'axeb = a'axb$, how to proceed. – user120386 Nov 03 '16 at 06:09
  • @ user120386 $b'b = a'axb = (yb) x b = y(bx)b = y e b = y b = a'a$. – bing Nov 03 '16 at 07:13
  • @ bing : Related to this problem if $(a,b) \in \mathcal H$ iff there exist inverse $a' , b'$ of $a$ and $b$ such that $a'a = b' b$ and $aa' = bb'$ – Struggler Nov 04 '16 at 03:16
  • $(a,b) \in \mathcal{R}$, we have $aa'=bb'$, and you refer to Howie's book "Fundamentals of semigroup Theory ". – bing Nov 04 '16 at 08:56
  • @ bing : in the case of $(a,b) \in \mathcal R$ , the inverse of $a$ and $b$ are different from $a'$ and $b'$ which are exist in the question . – Struggler Nov 04 '16 at 15:27
  • In the book which you refer me Howie used the theorem 2.3.4 (2) says if $b \in \mathcal D_a$ such that $\mathcal R_a \cap \mathcal L_b$ and $\mathcal R_b \cap \mathcal L_a$ contains an idempotent $e$ and $f$ respectively then $\mathcal H_b = \mathcal L_e \cap \mathcal L_f$ contains an inverse $a^$ of $a$ such that $a^ a = f$ and $aa^*= e$. Since $a'a \in \mathcal L_a = \mathcal L _b$ and $aa' \in \mathcal R_a = \mathcal R _b$ and both are idempotents. how to show that $a'a \in \mathcal R_a = \mathcal R_b$ and $aa' \in \mathcal L_a = \mathcal L_b$ – Struggler Nov 04 '16 at 15:38
  • I think we can use this result if $e,f$ are idempotent in a semigroup $S$ , then for all $x \in \mathcal L_f \cap \mathcal R_e$ there exist $y \in \mathcal R_f \cap \mathcal L_e$ such that $xy = e $ and $yx = f$. Here $e = aa'$ and $f =aa'$, and $a'a \in \mathcal L_a , aa' \in \mathcal R_a$ , thus $a \in \mathcal L_{a'a} \cap \mathcal R_{aa'}$ so there exist $b' \in \mathcal R_{a'a} \cap \mathcal L_{aa'}$ such that $bb' = aa'$ and $b'b = a'a$ – Struggler Nov 04 '16 at 16:12
  • Yes. It is shown in Proposition 2.4.1 in Howie' book. – bing Nov 05 '16 at 02:34