## The semigroup of nonempty subsets of a semigroup

Let $S$ be a semigroup, and let $P(S)$ denote the set of all nonempty subsets of $S$. Show that $P(S)$ is a semigroup under the operation $AB = \{ab \ |\ a \in A, b \in B\}$. Which of the basic properties of $S$ are preserved by $P(S)$?

We need only show that this operator is associative. Indeed, $(AB)C = \{ab\ |\ a \in A, b \in B\}C$ $= \{(ab)c\ |\ a \in A, b \in B, c \in C\}$ $= \{a(bc)\ |\ a \in A, b \in B, c \in C\}$ $= A\{bc\ |\ b \in B, c \in C\}$ $= A(BC)$.

Suppose $S$ has a left identity $e$. Then for all $A \in P(S)$, we have $\{e\}A = \{ea \ |\ a \in A\}$ $= \{a \ |\ a \in A\}$ $= A$. That is, $\{e\}$ is a left identity in $P(S)$. Similarly if $e$ is a right identity in $S$, then $\{e\}$ is a right identity in $P(S)$. Then if $e$ is an identity in $S$, $\{e\}$ is an identity in $P(S)$.

Suppose $S$ has a left zero $z$. Then for all $A \in P(S)$, we have $\{z\}A = \{za\ |\ a \in A\}$ $= \{z\ |\ a \in A\}$ $= \{z\}$. Thus $\{z\}$ is a left zero in $P(S)$. Similarly, if $z$ is a right zero in $S$ then $\{z\}$ is a right zero in $P(S)$, and if $z$ is a zero in $S$ then $\{z\}$ is a zero in $P(S)$.

If $S$ is commutative, then for all $A,B \in P(S)$ we have $AB = \{ab \ |\ a \in A, b \in B\}$ $= \{ba \ |\ b \in B, a \in A\}$ $= BA$. Thus $P(S)$ is also commutative.

Suppose $e \in S$ is an identity element and $u \in S$ is a unit, with $uv = vu = e$. Then $\{u\}\{v\} = \{v\}\{u\} = \{e\}$, so that $\{u\}$ is a unit in $P(S)$.

Suppose $I \subseteq S$ is a left ideal. Now let $A \subseteq S$ and $B \subseteq I$ be nonempty subsets. Now $AB = \{ab \ |\ a \in A, b \in B\} \subseteq I$, since $B \subseteq I$, and moreover this set is nonempty since $A$ and $B$ are nonempty. That is to say, $AB \in P(I)$. So $P(I) \subseteq P(S)$ is a left ideal. Likewise if $I$ is a right ideal in $S$, then $P(I)$ is a right ideal in $P(S)$, and if $I$ is a two-sided ideal in $S$, then $P(I)$ is a two-sided ideal in $P(S)$. More generally, if $T \subseteq S$ is a subsemigroup, then $P(T) \subseteq P(S)$ is a subsemigroup.

Suppose $S$ is a left zero semigroup. That is, $ab = a$ for all $a,b \in S$. If $A,B \in P(S)$, then $AB = \{ab \ |\ a \in A, b \in B\}$ $= \{a \ |\ a \in A, b \in B\}$ $= A$. That is, $P(S)$ is also a left zero semigroup. Similarly, if $S$ is a right zero semigroup then so is $P(S)$. Suppose now that $S$ has a zero element $0$, and that $S$ is a zero semigroup. Now $\{0\}$ is a zero in $P(S)$, and for all $A,B \in P(S)$, we have $AB = \{ab \ |\ a \in A, b \in B\}$ $= \{0 \ |\ a \in A, b \in B\}$ $= \{0\}$. So $P(S)$ is also a zero semigroup.

If $S$ is simple, $P(S)$ need not be simple, as we show. Consider the group $G = Z_2 = \{1,x\}$. We can easily verify that the only two-sided ideal in $G$ is $G$ itself, so that $G$ is simple as a semigroup. Now $P(Z_2) = \{\{1\}, \{x\}, Z_2\}$, and evidently, $\{Z_2\}$ is an ideal in $P(Z_2)$, so that $P(Z_2)$ is not simple.