# Question on regular group actions

1. Sep 2, 2012

### mnb96

Hello,
we know that if $\bullet : G\times A \rightarrow A$ is a regular action of G on the set A, then the action $\bullet$ is isomorphic to the action of G on itself, where the action on G is given by the group operation, that is: $\circ : G\times G \rightarrow G$ is defined as $g\circ g'=gg'$.

My question is: if we have a regular action of G on A, and a isomorphism $f:A\rightarrow G$ of actions, can we find a binary operation * for the set A, such that (A,*) is a group isomorphic to G?

***** EDIT: ******
My attempt to a possible solution was to define: $* : A\times A \rightarrow A$ as follows: $$a*a' = f(f(a) \bullet a')$$ which yields:

$$f(a) \circ f(a') = f(a)f(a') = gg'$$

Last edited: Sep 2, 2012