Let $(S,\cdot)$ be a semigroup. Then $x \in S$ is regular if there exists $y \in S$ such that $xyx=x$. A semigroup $S$ is regular if for all $x\in S,$ $x$ \is regular.
If $S$ is a semigroup and $x,y \in S$ we have $$(x,y) \in \mathcal{L} \iff (\exists p,q \in S^1)(x=py) \land (y=qx). $$ $$(x,y) \in \mathcal{R} \iff (\exists p,q \in S^1)(x=yp) \land (y=xq). $$ $$(x,y) \in \mathcal{H} \iff (x,y) \in \mathcal{L}\cap \mathcal{R}. $$ $$(x,y) \in \mathcal{D} \iff (x,y) \in \mathcal{L}\circ \mathcal{R}. $$ $$(x,y) \in \mathcal{J} \iff (\exists p,q,r,s \in S^1)(x=pyq) \land (y=rxs). $$
If $x \in S$ we define
$$V(x)= \{y \in S: (xyx=x) \land (yxy=y)\}.$$
I am reading Howie's book, more precisely the following propositions:
Theorem 2.3.4. Let a be an element of a regular $\mathcal{D}$-class D in a semigroup $S.$
- If $a' \in V(a),$ then $a' \in D$ and the two $\mathcal{H}$-classes $R_a \cap L_{a'}, L_a \cap R_{a'}$ contain, respectively, the idempotents $aa'$ and $a'a.$
- If $b\in D$ is such that $R_a \cap L_b$ and $L_a \cap R_b$ contain idempotents $e,f$ respectively, then $H_b$ contains an inverse $a^*$ of $a$ such that $aa^* = e, a^*a=f.$
I have problems to understand the proof of the following proposition.
Proposition 2.4.1. Let $a,b \in S$ be a regular semigroup. Then $(a,b) \in \mathcal{L}$ if and only if there exist $a' \in V(a)$ and $b' \in V(b)$ such that $a'a=b'b$. But I can't understand the proof of the book, more precisely the fact that it uses Theorem 2.3.4. (in red at the next image).
[Proof of the book][1] [1]: https://i.sstatic.net/wiNTq7LY.png
So here is what I have for the proof:
If $(a,b) \in \mathcal{L}.$ Then $L_a = L_b.$ Note that if $a' \in V(a)$ then $a'a \in L_a=L_b$. Moreover, $a'a$ is idempotent.
Now, we have that $b$ \is regular, then $D_b$ is regular so exist $e \in R_b$ such that $e$ is idempotent. Then $R_e=R_b.$
For $e$ idempotent, we have that for all $z \in R_e$ $ez=z.$ So $eb=b$ then $eb \in L_b \cap R_b.$
