Let $S$ be a semigroup and $a \in S$. An element $a' \in S$ is called an inverse of $a$ if $$ aa'a = a \qquad a'aa' = a'. $$ Denote the set of all inverses by $V(a)$. A semigroup where every element has an inverse is called a regular semigroup. We use Green's relations, in particular the definition of $\mathcal L-, \mathcal R-, \mathcal D-$ and $\mathcal H$-classes.
The following is taken from J. M. Howie Fundamental of semigroup theory (page 49):
Let $a,b$ be elements of a regular semigroup. Then $(a,b) \in\mathcal L$ if and only if there exists $a' \in V(a), b' \in V(b)$ such that $a'a = b'b$.
I tried to prove it on my own, but failed. And I do not understand the proof in the book. So can anyone give me proof of it or clarify the one given in the book? By the way, I was unable to find this proposition for regular semigroups in other books.
The proof from the book uses two previously shown Propositions, namely
In a regular semigroup (or in a regular $\mathcal D$-class, i.e. every element is regular) each $\mathcal L$-class and each $\mathcal R$-class contains an idempotent.
If $D$ is a regular $\mathcal D$-class and $a,b \in D$ are such that $R_a \cap L_b$ and $L_a \cap R_b$ contain idempotents $e,f$ respectively, then the $\mathcal H$-class of $b$ contains an inverse $a^*$ of $a$ such that $aa^* = e, a^*a = f$.
Now the proof goes like this:
Suppose that $(a,b) \in \mathcal L$. If $a' \in V(a)$ then $a'a$ is an idempotent in $L_a = L_b$. The $\mathcal R$-class $R_b$ contains at least one idempotent by Proposition 1 above, and then, by Proposition 2 above, the $\mathcal H$-class $R_{a'a} \cap R_e$ contain an inverse $b'$ of $b$ with the property that $b'b = a'a$ (and $bb' = e$). Notice that we have shown the stronger implication that $$ (a,b) \in \mathcal L \Rightarrow (\forall a' \in V(a))(\exists b' \in V(b)) a'a = b'b. $$
That's it. I found that in an older book by the same author, named Introduction to semigroup theory (which seems to be a precursor of the present book) the same Proposition appears, but the referred $\mathcal H$-class is $R_{a'a} \cap L_e$ and not $R_{a'a} \cap R_e$ (which seems to make more sense as $\mathcal H$-classes are intersection of $\mathcal R$-classes and $\mathcal L$-classes by definition, but note that the $\mathcal H$-class of $b$ is $L_b \cap R_b = L_{a'a} \cap R_e$).
But what I do not understand is that to apply Proposition 2 above, we need something like $L_a \cap R_b$ and $R_a \cap L_a$ should contain idempoents, but I do not see that $a'a$ is in $R_a$ (or $R_b$) neither do I see that $e$ is in $L_a = L_b$, and even if I write $e \in R_b \cap L_e$ then I need $a'a \in R_e \cap L_a$, but this I neither see?
EDIT: I found the following related question here.
