Action of Group on Coset Space is Group Action

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

Let $$X$$ be the set of all subsets of $$\left({G, \circ}\right)$$.

For any $$S \in X$$, we define $$\forall g \in G, S \in X: g \wedge S = g \circ S$$

This is a group action.

The following results hold about the stabilizers of this group action:


 * 1) $$\forall S \in X: \left|{\operatorname{Stab} \left({S}\right)}\right| \le \left|{S}\right|$$;
 * 2) $$\forall S \in X: \forall s \in S: \operatorname{Stab} \left({S}\right) \circ s \subseteq S$$;
 * 3) $$\forall S \in X: S \le G \implies \operatorname{Stab} \left({S}\right) = S$$.

When $$H \le G$$ we have $$\operatorname{Orb} \left({H}\right) = G / H$$, the left coset space of $$G$$ modulo $$H$$.

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

Let $$g \in G, s \in S$$.


 * Since $$G$$ is closed, and $$g S$$ consists of products of elements of $$G$$, it follows that $$g \wedge S \subseteq G$$.
 * $$e \wedge S = e S = \left\{{e s: s \in S}\right\} = \left\{{s: s \in S}\right\} = S$$.


 * Let $$g, h \in G$$. We have:

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


 * Let $$S \in X \implies S \subseteq G$$.

$$ $$ $$ $$


 * Now, let $$H \le G$$. Then:

$$ $$


 * $$\operatorname{Orb} \left({H}\right) = G / H$$ follows directly from the definitions.