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Let $S$ be a semigroup. The power semigroup of $S$ is the set $P(S)=2^S\setminus\lbrace\varnothing\rbrace $ with the operation $$AB=\lbrace ab\ |\ a\in A,\ b\in B\rbrace.$$

This operation is associative so the algebraic structure is a semigroup.

This semigroup can have a zero element. I've found the following examples.

$(a)$ If $S$ has a zero element $0$, then $\lbrace0\rbrace$ is the zero element of $P(S)$.

$(b)$ If $G$ is a group, then $G$ is the zero element of $P(G).$ Conversely, if $S$ is a semigroup, and $S$ is the zero element of $P(S)$, then $S$ is a group.

$(c)$ If $G$ is a group, then add a new identity element $1_1$ to $G$ to obtain a monoid $G_1=G\cup\lbrace 1_1\rbrace$ in which the multiplication is left unchanged on $G$ and the rest of products are defined by $1_1x=x1_1=x.$ Then $G$ is the zero element of $P(G_1)$. We can then define $G_n$ to be the monoid obtained by adding a new identity element $1_k$ to $G$ $n$ times (that is for $k=1,\ldots,n$). $G$ is also the zero element of $P(G_n).$

$(d)$ To obtain an example that doesn't have a zero and isn't a monoid, we can take $G_\omega=G\cup\lbrace1_1,1_2,\ldots\rbrace$ where $1_k$ are pairwise different and $1_k$ is the identity element of $G\cup\lbrace1_1,\ldots,1_{k}\rbrace.$ We can define such a semigroup $G_{\iota}$ for any ordinal number $\iota$ and always $G$ will be the zero element of $P(G_\iota).$

I suspect there are more examples, but I don't know how to find them. I would like to have a theorem of the form

$P(S)$ has a zero element iff $\phi(S)$. Then the zero element is $\psi(S)$.

where $\phi(S)$ is a simple and edifying formula taking a semigroup as its argument, and $\psi(S)$ is a function that takes a semigroup and gives a subset of the semigroup.

In the other direction, let's suppose $I\subseteq S$ is the zero element of $P(S).$ This means that for any $X\subseteq S$, we have $XI=IX=I.$ This is equivalent to $xI=Ix=I$ for any $x\in S.$ This condition, if I'm not mistaken, implies the following conditions.

$(1)$ $I$ is both the smallest left ideal of $S$ and the smallest right ideal of $S$.

$(2)$ $I$ is an $\mathscr L$-class of $S;$ $I$ is an $\mathscr R$-class of $S;$ $I$ is a $\mathscr J$-class of $S$.

$(3)$ $I$ is a subgroup of $S$.

I'm not sure if these conditions, in conjunction, suffice to make $I$ the zero element of $S$. I suspect not. Could you please make the picture clearer for me?

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1 Answer 1

up vote 5 down vote accepted

You are right. More simply: $I$ is a zero in $P(S)$ if and only if $I$ is an ideal of $S$ and a subgroup of $S$. Indeed, if $I$ is a zero, it must be a subgroup and an ideal as you wrote yourself. On the other hand, if $I$ is an ideal and a subgroup, then for every $x$ in $S$, $Ix$ is a left ideal of $S$ contained in $I$, hence it is a left ideal of $I$, so $Ix=I$ (a group has only one left ideal). Similarly, $xI=I$. So $I$ is a zero.

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Right, thanks a lot! But which semigroups have such an ideal? Are there any examples other than the ones I mentioned? Can we characterize such semigroups? –  Michał Masny Oct 23 '12 at 11:44
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Here is a general construction. Take any semigroup $T$, any group $G$ and any homomorphism $\phi: T\to G$. Consider $S$ the disjoint union of $T$ and $G$ with operation * that coincides with the native operations on $T,G$ and $t*g=\phi(t)g$, $g*t=g\phi(t)$ for every $g\in G$, $t\in T$. Then $S$ is a semigroup (the semilattice of $T$ and $G$) and $G$ is an ideal of $S$. These are not all semigroups with subgroups which are ideals, but it gives a lot of examples. –  Mark Sapir Oct 23 '12 at 11:50
    
Your examples are like that where $T$ is the chain of idempotents ($\{e_1,...,e_n\}$, $e_ie_j=e_je_i=e_i$ if $i\le j$),and $\phi$ is a homomorphism that maps everything to 1. –  Mark Sapir Oct 23 '12 at 11:56
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I think that looking for more and more complicated examples is not the way to prove what you want. I think you know enough to be able to prove it (or find a relatively easy counterexample). –  Mark Sapir Oct 23 '12 at 14:07
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Note if S is a finite monoid with group of units H, then P(S) has group of units H. So if P(S) is isomorphic to P(G) then G is isomorphic to H and for cardinality reasons S=G. So the question is easy if G is finite. –  Benjamin Steinberg Oct 24 '12 at 3:09

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