# When does a power semigroup have a zero, and what can the zero be?

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?

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.
• 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. – user6976 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. – user6976 Oct 23 '12 at 11:56