Conjugacy Action on Subgroups is Group Action

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

Let $X$ be the set of all subgroups of $G$.

For any $H \le G$ and for any $g \in G$, we define: $\forall g \in G, H \in X: g * H = g \circ H \circ g^{-1}$

This is a group action.

Also:
 * $\operatorname{Stab} \left({H}\right) = N_G \left({H}\right)$ where $N_G \left({H}\right)$ is the normalizer of $H$ in $G$.


 * $\operatorname{Orb} \left({H}\right)$ is the set of subgroups conjugate to $H$.

Proof
Clearly GA-1 is fulfilled as $e * H = H$.

GA-2 is shown to be fulfilled thus:

$\operatorname{Stab} \left({H}\right) = \left\{{g \in G: g \circ H \circ g^{-1} = H}\right\}$ which is how the normalizer is defined.

$\operatorname{Orb} \left({H}\right) = \left\{{g \circ H \circ g^{-1}: g \in G}\right\}$ from the definition.