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?

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?