Conjugacy Action on Group Elements is Group Action

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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

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