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

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?

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

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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 $$b \not\mathrel{\cal H} ba$$.