Anyone knows how to describe explicitly the submonoids of Z_n, regarded as a multiplicative monoid?
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2$\begingroup$ Martin, I don't understand your hint. $Z_n$ (which I assume to be integers modulo $n$) is not a group wrt multiplication, and unless $n$ is prime contains non-zero nilpotents, so is not even a sum of submonoids of groups $\endgroup$– Yemon ChoiMar 24, 2012 at 20:53
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$\begingroup$ Right, I've deleted my comment. Thanks. $\endgroup$– Martin BrandenburgMar 24, 2012 at 21:31
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2 Answers
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Subsemigroups of finite commutative (even cyclic) semigroups are complicated. For example every finite lattice embeds into the lattice of subsemigroups of a finite cyclic semigroup (hence into the lattice of submonoids of one of the $\mathbb{Z}_n$. See Repnitskii, Vladimir, On subsemigroup lattices without nontrivial identities. Algebra Universalis 31 (1994), no. 2, 256–265.
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To supplement Mark's answer, note that every non-invertible element of $Z/p^n$ is nilpotent. So the idempotents in this case are 0,1. Thus if m is divisible by exactly k primes, then $Z/m$ has semilattice of idempotents the power set of a k-element set by the Chinese remainder theorem. Of course every k element semilattice embeds in the power set of a k-element set so even the lattice of semilattice submonoids can be complicated.
answered Mar 25, 2012 at 13:11