Right Regular Representation by Inverse is Transitive Group Action

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Theorem

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

Let $*: G \times G \to G$ be the group action:

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

where $\rho_g$ is the right regular representation of $G$ with respect to $g$.


Then $*$ is a transitive group action.


Proof

Let $g, h \in G$.

Then:

\(\ds \exists a \in G: \, \) \(\ds h\) \(=\) \(\ds a \circ g^{-1}\) Group has Latin Square Property
\(\ds \leadsto \ \ \) \(\ds h\) \(=\) \(\ds \map {\rho_{g^{-1} } } a\) Definition of Right Regular Representation
\(\ds \leadsto \ \ \) \(\ds h\) \(=\) \(\ds g * a\) Definition of $*$

$h$ is arbitrary, therefore the above holds for all $h \in G$.

By definition of orbit:

$\Orb h = \set {t \in G: \exists g \in G: g * h = t}$

That is:

$\Orb h = G$

Hence the result by definition of transitive group action.

$\blacksquare$


Also see


Sources