Conjugacy Action on Group Elements is Group Action/Proof 2

Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

The conjugacy action on $G$:

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

is a group action on itself.

Proof

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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:

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

is a group action, as required.

$\blacksquare$