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we know that if [itex]\bullet : G\times A \rightarrow A[/itex] is aregularaction of G on the set A, then the action [itex]\bullet[/itex] is isomorphic to the action of G on itself, where the action on G is given by the group operation, that is: [itex]\circ : G\times G \rightarrow G[/itex] is defined as [itex]g\circ g'=gg'[/itex].

My question is: if we have a regular action of G on A, and a isomorphism [itex]f:A\rightarrow G[/itex] 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: [itex]* : A\times A \rightarrow A[/itex] as follows: [tex]a*a' = f(f(a) \bullet a')[/tex] which yields:

[tex]f(a) \circ f(a') = f(a)f(a') = gg'[/tex]

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# Question on regular group actions

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