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A semigroup $(A,\cdot)$ that is idempotent (i.e. $a^2=a$ for every element $a\in A$) is naturally generated by its subgroups (every element on itself constitutes a trivial group). I would like to know what is known about semigroups that are generated by (cyclic) groups? Again, such groups do not need to share a common neutral element. I am interested in the commutative case mainly, but I appreciate any comments.

In particular, I would like to know references to this topic, since I haven't found anything so far. Thank you.

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  • $\begingroup$ You asked this question today on the group-pub-forum, and it has been answered there by Markus Pfeiffer -- what do you miss in his answer? $\endgroup$
    – Stefan Kohl ♦
    Apr 15, 2016 at 13:30
  • $\begingroup$ Thanks for notice! I did not expect such a prompt response. You guys are really fast :). I have still to think about Markus Pfeiffer's answer. $\endgroup$
    – Miroslav Korbelar
    Apr 15, 2016 at 13:40

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In the non-commutative case you can't say much. The monoid of all maps on n letters is generated by the symmetric group and an idempotent. In fact semigroups generated by idempotents can be quite wild. The singular nxn matrices over a field are generated by idempotents.

In the commutative case much more can be said. In the finite case your semigroup will be a commutative inverse semigroup and so be a semilattice of groups (also called a Clifford semigroup). I guess this should also be true in the infinite case too.

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