Doubt in Location Lemma in Greens Relation Abstract Algebra!

Question

I am unable to prove one part of rectangular lemma in green's relations. Let $S^1$ be a monoid. Then I need to prove that $m.m' \in D(m) \iff m.m'\in R(m) \cap L(m')$. How should I go about proving this?
$R(m)$ is Right equivalence classes of m.
$L(m')$ is Left equivalence classes of m'.
$D(m)$ is the equivalence class of m, defined by relation $D=L\circ R=R\circ L$.
I know that $x\in R(m) \iff \exists a,b:xa=m$ and $x=mb$. Symmetrically for $L(m')$.
Also, $D=L\circ R=R\circ L$ is an equivalence relation.

Answer 1

This is not true in arbitrary monoids. For instance, it is not true in the bicyclic monoid.

The bicyclic monoid is the monoid with presentation $\langle \{a, b\} \mid ab = 1 \rangle$. I let you verify that every element of the bicyclic monoid can be written uniquely in the form $b^ma^n$, where $m, n \geq 0$. The product of such words is given by $$ (b^ma^n)(b^pa^q) = b^ra^s $$ where $r = m - n + \max(n,p) = m + p - \min(n,p)$ and $s = q - p + \max(n,p) = n + q - \min(n,p)$. The idempotents are the elements of the form $b^na^n$, with $n \geqslant 0$. It contains a single $\cal D$-class and $\cal D = \cal J$. Its egg-box picture is represented below:

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Let $R(x)$, $L(x)$ and $D(x)$ be the $\cal R$-class, the $\cal L$-class and the $\cal D$-class of an element $x$. Then $b, ba \in D(b)$. It is also true that $bba \in L(ba)$. However, $bba \notin R(b)$.