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Show the only (non-trivial) subdirectly irreducible right-zero semi-group is the 2-element one. (I only need to restrict myself to the finitely generated ones).

A right-zero semi group is a semi-group + $xy=y$. Let $RZ$ be a finitely generated right-zero semi-group

If $Con(RZ) - \{\Delta\}$ has a minimum element then $RZ$ is subdirectly irreducible. It is clear the the two element rz-semi-group is subdirectly irreducible as it only has two congrences, namely $\nabla, \Delta$.

I am having trouble using this to get that for an rz-semi-group with three or more elements it will not be subdirectly irreducible. Maybe separating of points would be useful but I haven't been able to get it to work.

oliverjones
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1 Answers1

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Hint. Let $Z_3 = \{a_1, a_2, a_3\}$ be the three-element right-zero semigroup. Show that $a_1 \sim_1 a_2$ and $a_2 \sim_2 a_3$ define two congruences $\sim_1$ and $\sim_2$ and that $Z_3/{\sim_1} = Z_3/{\sim_2} = Z_2$.

J.-E. Pin
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    Or put in another way, if $\mathbf{S} = \langle S, \cdot \rangle$ is a right-zero semigroup (or left-zero semigroup, or null semigroup, among others, I suppose...), then every equivalence relation on $S$ is a congruence on $\mathbf{S}$. Thus $\mathbf{Con,S} = \Pi(S)$, the lattice of partitions of $S$. And a lattice of partitions has a single atom iff it is the lattice of partitions on a two-element set. – amrsa Nov 01 '16 at 10:00