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A semigroup is called globally idempotent when for any $x\in S$ there are $y,z\in S$ such that $x=yz$.

Is there a name for monoids whose every globally idempotent subsemigroup contains the identity element?

For example, the monoid $(\mathbb N,+)$ has this property, because if $S$ is a globally idempotent subsemigroup of $\mathbb N$, and $a$ is the smallest element of $S$, then $2a$ is the smallest element of $S+S$. Therefore $a=2a$, so $a=0$.

But there are many non-examples, even among groups.

Is anything known about such monoids?

Added: I'll explain why I'm asking about these. If $M$ is a monoid whose every globally idempotent subsemigroup contains the identity, then the natural order $\leq$ on idempotents of the power semigroup $P(M)$ coincides with $\supseteq$. And this seems like a very nice property.

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  • $\begingroup$ Probably torsion, since every one generated subsemigroup which is globally idempotent is finite. Of course, the situation isn't quite that simplistic, but that seems to be a good starting point. Gerhard "Ask Me About System Design" Paseman, 2012.10.29 $\endgroup$
    – Gerhard Paseman
    Commented Oct 29, 2012 at 20:15
  • $\begingroup$ @Gerhard I don't understand. Isn't $(\mathbb N\setminus\lbrace0\rbrace,+)$ both one generated and globally idemopotent? In fact, $(\mathbb Z,+)$ isn't torsion, and if $S$ is its globally idempotent subsemigroup, then if all elements of $S$ have the same sign, $S$ contains $0$, by the same argument as for $\mathbb N$. If there are $x,y\in S$ with $xy<0$ then $0=|y|x+|x|y\in S.$ So $\mathbb Z$ also has this property. $\endgroup$
    – Michał Masny
    Commented Oct 29, 2012 at 21:05
  • $\begingroup$ In your first example, what two elements add to 1? Gerhard "Ask Me About Adding Ones" Paseman, 2012.10.29 $\endgroup$
    – Gerhard Paseman
    Commented Oct 29, 2012 at 22:08
  • $\begingroup$ @Gerhard Oh, sorry. You're right. Something went wrong in my brain. But still, the second example proves they don't have to be torsion, right? (Sorry, all my confidence has just disappeared.) $\endgroup$
    – Michał Masny
    Commented Oct 29, 2012 at 22:41
  • $\begingroup$ As a semigroup under addition, Z is not one generated. And it is true that some of what you are interested in lacks torsion. However, if none of the one generated subsemigroups is globally idempotent, then you have no torsion at all, and that may tell you something about your picture. For general semigroups, I don't know what the answer is, but for commutative finitely generated examples, you may have something like a classification of f.g. Abelian groups possible. That last is a guess, but it is worthy of consideration. Gerhard "Also, Could Be Globally Wrong" Paseman, 2012.10.29 $\endgroup$
    – Gerhard Paseman
    Commented Oct 29, 2012 at 23:23

2 Answers 2

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There is of course no such name, as this wouldn't bear much information about the semigroup.

Now, some beautiful paper, where the global idempotency really appears, and really means something:

Robertson, E. F.; Ruškuc, N.; Wiegold, J. Generators and relations of direct products of semigroups. Trans. Amer. Math. Soc. 350 (1998), no. 7, 2665–2685.

Also, if you are interested in power semigroups or finitary power semigroups, there is a whole thesis by Peter Gallagher (with very nice results about finitary power semigroups of groups, and some nice chacracterisation theorems and some really difficult to come up with examples!) from St Andrews.

Also, there is a series of papers by Volodymyr Mazorchuk on various power semigroups of finite transformation semigroups. Also check on the web some results by Mykola Rybak (I think from the journal "Algebra and Discrete Mathematics")

There is some work by Ash -- which is great if you like finite semigroups, you can find a lot about this in the recent book by Benjamin.

There is the most important result in this topic that there are two infinite non-isomorphic semigroups S and T, such that their finitary power semigroups are isomorphic.

Finally, the most important question in the topic is whether there are two non-isomorphic FINITE semigroups whose power semigroups would be isomorphic.

Also, just in case, my e-mail is: [email protected]

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  • $\begingroup$ Nice answer Victor. $\endgroup$
    – Benjamin Steinberg
    Commented Nov 2, 2012 at 13:17
  • $\begingroup$ Where could I read about the isomorphism problem for finite semigroups? $\endgroup$
    – Michał Masny
    Commented Nov 5, 2012 at 23:35
  • $\begingroup$ What about a reference for the existence of two infinite non-isomorphic semigroups whose finitary power semigroups are isomorphic? I was not aware of this result. $\endgroup$
    – Salvo Tringali
    Commented Oct 29 at 2:50
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I don't know if there is a name for this. Note that this can never happen for a finite monoid which is not a group. The minimal ideal of a finite semigroup is globally idempotent and if the finite monoid is not a group, this minimal ideal does not contain the identity.

Afterthought. Your semigroup cannot have any non-identity idempotents if its globally idempotent subsemigroups contain the identity.

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