Right Regular Representation by Inverse is Group Action

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $*: G \times G \to G$ be the operation:
 * $\forall g, h \in G: g * h = \rho_{g^{-1}} \left({h}\right)$

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

Then $*$ is a group action.

Proof
The group action axioms are investigated in turn.

Let $g, h, a \in G$.

Thus:

demonstrating that group action axiom $GA1$ holds.

Then:

demonstrating that group action axiom $GA2$ holds.

Also defined as
Some sources denote a mapping by placing the symbol defining that mapping on the left of its operand.

Thus under such a convention:
 * $\rho_g \left({a}\right)$ is written $a \rho_g$

and:
 * $g * a$ is written $a * g$ (or even $a g$ in sources which do not place a high regard on clarity).

Thus the right regular representation itself can directly be defined as being a group action without the need to take the inverse:

Such a treatment can be found in.

As it runs contrary to the conventions used on, beyond its mention here it will not be used.

However, this example indicates how the arbitrary nature of notational conventions can cause the details of results to be equally arbitrarily dependent upon the convention used.

Also see

 * Left Regular Representation is Group Action


 * Group Acts on Itself