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Let $E$ be a finite semigroup. According to N. Bourbaki (Algèbre I p. 121 exerc. 14 c), if $M$ and $M'$ are minimal right ideals in $E$, then they are isomorphic. I spent some time browsing through Clifford and Preston's monography "the algebraic theory of semigroups", also reading a few articles by Rees, Clifford and Dubreil, and found absolutely no mention of this fact anywhere. Maybe I haven't looked hard enough. If there is any semigroup specialist, could you tell me what you think?

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up vote 2 down vote accepted

This is in Clifford-Preston. First every minimal left (right) ideal is inside the minimal two-sided ideal $I$ (which is unique), see Exercise 13 on page 84. Second, the ideal $I$ is a simple semigroup (obvious, but is also in C-P). Third, by Sushkevich's theorem (Appendix A), all maximal subgroups of $I$ are isomorphic. By Theorem 1.27, every minimal left ideal $L$ in $I$ is a left group, i.e. the direct product of the maximal subgroup and a left zero semigroup (whose order does not depend of $L$). Hence all minimal left ideals are isomorphic.

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Got it. Thanks. – Pnine Jul 4 '12 at 7:40

I will use Green's relations, definitions are here if you don't know them :'s_relations.

If you represent the smallest D-class $C_\min$ by a box as in the wikipedia article, then the right (resp. left) minimal ideals are exactly the rows (resp. columns) of $C_\min$, i.e. its R- and L-classes. This can be seen by the fact that every right ideal must contain a R-class of $C_\min$ (otherwise it would not be an ideal), and that it must also be included in a single R-class of $C_\min$ to be minimal. Then it suffices to see that for all $e\in E$, if $R$ is a R-class of $C_\min$ , then so is $eR$, and $x\mapsto ex$ is an isomorphism between $R$ and $eR$.

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Thanks for your answer. Why should $x \mapsto ex$ be an isomorphism? This map is obviously one-one and onto but why do we have $exy=exey$ if $x,y \in R$? – Pnine Jul 3 '12 at 20:16

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