Conjugacy Action on Group Elements is Group Action/Proof 2

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

By definition, the (left) conjugacy action on subgroups is the group action $*_X : G \times X \to X$ defined as:
 * $g *_X X = g \circ X \circ g^{-1}$

By Conjugacy Action on Subgroups is Group Action, the (left) conjugacy action on subgroups $*_X$ is a group action.

By Subset Product Action is Group Action, it follows that the conjugacy action $*: G \times G \to G$ such that:


 * $\forall g, h \in G: g * h = g \circ h \circ g^{-1}$

is a group action, as required.