Subset Product: Powerset Semigroup

Theorem

Let $\left({G, \circ}\right)$ be a group.

Then its power set $\mathcal P \left({G}\right)$ is a semigroup $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ with respect to the subset product.

Proof

We need to prove closure and associativity.

Closure

Let $\left({G, \circ}\right)$ be a group, and let $A, B \subseteq G$.

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle \forall a \in A, b \in B: a \circ b \in G\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle A \circ B \subseteq G\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Subset Product
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle A \circ B \in \mathcal P \left({G}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Power Set

$\Box$

Associativity

This follows from Subset Product of Associative is Associative.

$\Box$

Thus $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ is a semigroup.

Sources

• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 41.1$