Conjugacy Action on Group Elements is Group Action

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

We have that:

$e * x = e \circ x \circ e^{-1} = x$

and so Group Action Axiom $\text {GA} 2$ is fulfilled.

Group Action Axiom $\text {GA} 1$ is shown to be fulfilled thus:

$\ds \paren {g_1 \circ g_2} * x$

$=$

$\ds \paren {g_1 \circ g_2} \circ x \circ \paren {g_1 \circ g_2}^{-1}$

Definition of $*$

$\ds $

$=$

$\ds g_1 \circ g_2 \circ x \circ g_2^{-1} \circ g_1^{-1}$

Inverse of Group Product

$\ds $

$=$

$\ds g_1 * \paren {g_2 \circ x \circ g_2^{-1} }$

Definition of $*$

$\ds $

$=$

$\ds g_1 * \paren {g_2 * x}$

Definition of $*$

$\blacksquare$

Proof 2

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.

$\blacksquare$

Also see

Sources

1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 5.5$. Groups acting on sets: Example $103$
1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.3$: Group actions and coset decompositions: Examples of group actions: $\text{(v)}$
1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.5$