How to prove that in a semigroup, two $\mathcal{L}$ classes in a $\mathcal{D}$ class are incomparable by $\leq_{\mathcal{L}}$ relation?

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Let $S$ be a semigroup. I would like to prove that two $L$ classes (namely $L_1$ and $L_2$) such that both are in a $D$ class (that is $L_1\subseteq D$ and $L_2 \subseteq D$) are incomparable by $\leq_{L}$ relation.
I am interested in the proof in cases when $S$ is
(a) infinite
(b) finite.

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Short answer. This property holds in any stable semigroup and hence in particular in any finite semigroup and in any compact semigroup.

It does not hold for instance in the Baer-Levi semigroup.

Definition.

A $\cal J$-class $J$ is left [right] stable if, for every $s, t \in J$, $s \leqslant_{\cal L} t$ implies that $s \mathrel{\cal L} t$ [$s \leqslant_{\cal R} t$ implies that $s \mathrel{\cal R} t$]. A $\cal J$-class $J$ is stable if it is both right and left stable. Finally a semigroup is stable if each of its $\cal J$-classes is stable.

In a stable semigroup, the Green's relations $\cal J$ and $\cal D$ are equal. Furthermore, the following properties hold:

  1. if $s = usv$ for some $u, v\in S^1$, then $us \mathrel{\cal H} s \mathrel{\cal H} sv$;
  2. If $s \mathrel{\cal J} t$ and $s \leqslant_{\cal R} t$ $[s \leqslant_{\cal L} t]$, then $s \mathrel{\cal R} t$ $[s \mathrel{\cal L} t]$;
  3. If $s \leqslant_{\cal J} sx$, then $s \mathrel{\cal J} sx$ and if $s \mathrel{\cal J} sx$, then $s \mathrel{\cal R} sx$.
  4. If $s \leqslant_{\cal J} xs$, then $s \mathrel{\cal J} xs$ and if $s \mathrel{\cal J} xs$ then $s \mathrel{\cal L} xs$.
  5. A $\mathrel{\cal D}$-class $D$ is regular if and only if there exist two elements of $D$ whose product belongs to $D$.