User:Austrodata/Left Group Action and Right Group Action are Equivalent
Jump to navigation
Jump to search
Let $G$ a group with $e$ as identity element.
Let $X$ a set.
Let $\lambda: G \times X \to X$ be a left action on $X$.
Then, the mapping $\rho: X \times G: (x,g) \mapsto \map {\lambda} {g^{-1}, x}$ is a right action.
Proof
First, check that the mapping $\rho$ satisfies the axioms of right action:
| \(\text {(RGA $1$)}: \quad\) | \(\ds \map {\rho} {\map {\rho} {x,g}, h}\) | \(=\) | \(\ds \map {\lambda} {h^{-1}, \map {\lambda} {g^{-1}, x} }\) | definition of $\rho$ | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \) | \(=\) | \(\ds \map {\lambda} {h^{-1} g^{-1}, x}\) | left group action axiom | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \) | \(=\) | \(\ds \map {\lambda} {\paren{gh}^{-1}, x}\) | inverse of product | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \) | \(=\) | \(\ds \map {\rho} {x,gh}\) | definition of $\rho$ |
$\Box$ | |||||||||
| \(\text {(RGA $2$)}: \quad\) | \(\ds \map {\rho} {x, e}\) | \(=\) | \(\ds \map {\lambda} {e^{-1}, x}\) | definition of $\rho$ | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \) | \(=\) | \(\ds \map {\lambda} {e, x}\) | inverse of identity | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \) | \(=\) | \(\ds e\) | left group action axiom |
$\Box$ |
$\Box$
Furthermore, since these are all equalities, $\rho$ must be bijective.
Hence, to each left action corresponds a right action and reversely.
$\blacksquare$
Source
- 2003: David S. Dummit and Richard M. Foote: Abstract Algebra (3rd ed.) : Chapter $4$: Group Actions $\S4.3$: Groups Acting on Themselves by Conjugation — The Class Equation