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.

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