Definition:Subset Product

Let $$\left({S, \circ}\right)$$ be an algebraic structure.

We can define an operation on the power set $$\mathcal {P} \left({S}\right)$$ as follows:

$$\forall A, B \in \mathcal {P} \left({S}\right): A \circ_{\mathcal{P}} B = \left\{{x = a \circ b: a \in A, b \in B}\right\}$$

This is called this the operation induced on $$\mathcal {P} \left({S}\right)$$ by $$\circ$$.

It is usual to write $$A \circ B$$ for $$A \circ_{\mathcal {P}} B$$.

If $$A = \varnothing$$ or $$B = \varnothing$$, then $$A \circ B = \varnothing$$.

Some results
Let $$\left({S, \circ}\right)$$ be a groupoid.

Let $$T \subseteq S$$.

Then $$\left({T, \circ}\right)$$ is a groupoid iff $$T \circ T \subseteq T$$.

Proof

By definition, $$T \circ T = \left\{{x = a \circ b: a, b \in T}\right\}$$.

Let $$\left({T, \circ}\right)$$ be a groupoid.

Then $$T$$ is closed, that is, $$\forall x, y \in T: x \circ y \in T$$.

Thus $$x \circ y \in T \circ T \Longrightarrow x \circ y \in T$$.

Now suppose $$T \circ T \subseteq T$$.

Then $$x \circ y \in T \circ T \Longrightarrow x \circ y \in T$$.

Thus $$T$$ is closed, and therefore $$\left({T, \circ}\right)$$ is a groupoid.

Let $$\left({S, \circ}\right)$$ be a groupoid.

If $$\circ$$ is associative, then $$\circ_{\mathcal {P}}$$ is associative.

Proof

Let $$\left({S, \circ}\right)$$ be a groupoid in which $$\circ$$ is associative.

Let $$X, Y, Z \in \mathcal {P} \left({S}\right)$$.

Then:


 * $$X \circ_{\mathcal {P}} \left({Y \circ_{\mathcal {P}} Z}\right) = \left\{{x \circ \left({y \circ z}\right): x \in X, y \in Y, z \in Z}\right\}$$


 * $$\left({X \circ_{\mathcal {P}} Y}\right) \circ_{\mathcal {P}} Z = \left\{{\left({x \circ y}\right) \circ z: x \in X, y \in Y, z \in Z}\right\}$$

... from which follows that $$\circ_{\mathcal {P}}$$ is associative on $$\mathcal {P} \left({S}\right)$$.

Let $$\left({S, \circ}\right)$$ be a groupoid.

If $$\circ$$ is commutative, then $$\circ_{\mathcal {P}}$$ is commutative.

Proof

Let $$\left({S, \circ}\right)$$ be a groupoid in which $$\circ$$ is commutative.

Let $$X, Y \in \mathcal {P} \left({S}\right)$$.

Then:


 * $$X \circ_{\mathcal {P}} Y = \left\{{x \circ y: x \in X, y \in Y}\right\}$$


 * $$Y \circ_{\mathcal {P}} X = \left\{{y \circ x: x \in X, y \in Y}\right\}$$

... from which it follows that $$\circ_{\mathcal {P}}$$ is commutative on $$\mathcal {P} \left({S}\right)$$.