Subset Product Action is Group Action

Theorem
Let $$\left({G}\right)$$ 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:
 * $$\forall g \in G: g * S = g S$$

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

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 $$g S$$ consists of products of elements of $$G$$, it follows that $$g * S \subseteq G$$.

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

and so GA-2 is satisfied.

Now let $$g, h \in G$$.

We have:

$$ $$ $$ $$ $$ $$

and so GA-1 is satisfied.

Hence the result.