$\mathcal{L}$ Green relation vs $\mathcal{H}$ Green relation

Question

Let $S$ be a regular semigroup. The Green's equivalence relation $\mathcal{L}$ is defined on $S$ as follows. $$s\mathcal{L}t \mbox{ if and only if } Ss=St.$$ Similarly using cyclic right ideals of $S$, the Green's relation $\mathcal{R}$ is defined. Finally note that $\mathcal{H}=\mathcal{L}\cap \mathcal{R}$. It is a known fact that $\mathcal{L}$ is a right congruence on $S$ and $\mathcal{R}$ is a left congruence. However for some classes of semigroups (e.g. commutative semigroups) we know that $\mathcal{L}=\mathcal{R}=\mathcal{H}$ and they are congruence on $S$.

My question is the following.

Is there a regular semigroup $S$ such that the Green relation $\mathcal{L}$ on $S$ is a congruence but $\mathcal{H}$ is not a congruence?

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Sorry! I correct a mistake about completely simple semigroups in my question.

Answer 1

Let $M$ be the monoid presented by $\langle a, b \mid a^2 = 1, b^2 = b, ba = b\rangle$. Then $M = \{1, a, b, ab\}$ and its group of units is $\{1, a\}$. Thus $1 \mathrel{\cal H} a$ and hence $1 \mathrel{\cal L} a$. Moreover $b \mathrel{\cal L} ab$. The relation $\cal L$ is a congruence, but $\cal H$ is not, since $1 \mathrel{\cal H} a$ but $b \not\mathrel{\cal H} ba$.