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Are there examples of (finite) semigroups $S$ and $T$ such that they have the same 'number' of $\mathscr D$-classes, $S$ has idempotents and $T$ doesn't? Alternatively, they both may have idempotents, but one of them has multiple idempotents in one $\mathscr D$-class while the other may have at most one idempotent in each $\mathscr D$-class.

An obvious candidate would be a semigroup $T$ that is not a monoid and $S := T^1$ but then does the external identity constitute a new $\mathscr D$-class on its own or does it fit in somewhere?

AlvinL
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1 Answers1

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If you take a finite semigroup, it contains at least one idempotent. Thus your first question has no solution.

For your second question, take $S = \{a, 0\}$ with $aa = a0 = 0a = 0$ and $00 = 0$ and $T = \{1, 0\}$ with the usual multiplication of integers. Both semigroups have two $\mathcal{D}$-classes, but $0$ is the unique idempotent of $S$ while $T$ has two idempotents.

Finally, if you want to have the same number of idempotents but a different number of $\mathcal{D}$-classes, take $R = \{a, b\}$ with $aa = ba = a$ and $ab = bb = b$ and $T$. Both $R$ and $T$ have two elements, which are both idempotent. Now $a$ and $b$ are in the same $\mathcal{D}$-class of $S$ but $1$ and $0$ are in different $\mathcal{D}$-classes of $T$.

J.-E. Pin
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