Subset Product Action is Group Action

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $\mathcal P \left({G}\right)$ be the power set of $\left({G, \circ}\right)$.

For any $S \in \mathcal P \left({G}\right)$, we define:
 * $\forall g \in G: g * S = g \circ S$

where $g \circ S$ is the subset product $\left\{{g}\right\} \circ S$.

This is a group action.

Proof
The fact that this is a group action follows directly from the definitions:

Let $g \in G$.

First we note that since $G$ is closed, and $g \circ S$ consists of products of elements of $G$, it follows that $g * S \subseteq G$.

Next we note:
 * $e * S = e \circ S = \left\{{e \circ s: s \in S}\right\} = \left\{{s: s \in S}\right\} = S$

and so GA-2 is satisfied.

Now let $g, h \in G$.

We have:

and so GA-1 is satisfied.

Hence the result.