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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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  • $\begingroup$ Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$
    – Shaun
    Commented Mar 6, 2023 at 21:27

1 Answer 1

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

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  • $\begingroup$ Thanks so much dear professor for your nice example! By the way, do you know an up-to-date reference about Green's relations? $\endgroup$
    – khers
    Commented Mar 14, 2023 at 8:43

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