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.



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> $(a)$ If $S$ has a zero element $0$, then $\lbrace0\rbrace$ is the zero element of $P(S)$.
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> $(b)$ 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.
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> $(c)$ 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 $P(G\_n).$
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> $(d)$ 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 $P(G\_\iota).$
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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



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



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> $(1)$ $I$ is both the smallest left ideal of $S$ and the smallest right ideal of $S$.
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> $(2)$ $I$ is an $\mathscr L$-class of $S;$ $I$ is an $\mathscr R$-class of $S;$ $I$ is a $\mathscr J$-class of $S$.
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> $(3)$ $I$ is a subgroup of $S$.
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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?



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


