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
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.3$: Group actions and coset decompositions: Examples of group actions: $\text{(iv)}$