Does a Group Action Always Use the Group's Original Operation?

[Bachelier]

A group $G$ is said to act on a set $X$ when there is a map $\phi:G×X \rightarrow X$ such that the following conditions hold for any element $x \in X$.

1. $\phi(e,x)=x$ where $e$ is the identity element of $G$.

2. $\phi(g,\phi(h,x))=\phi(gh,x) \ \ \forall g,h \in G$.

My question is: is this action on the set $X$ performed under the operation of the group $G$ or under a different new operation. Only the $Wikipedia$ article author defines this operation as the group $G$ original operation. On the other hand, I was reading a different book and it defines the action using a totally new operation. Mind you this book is quite old.

 

[Stephen Tashi]

My question is: is this action on the set $X$ performed under the operation of the group $G$ or under a different new operation.

Are you asking whether the notation $gh$ refers to the multiplication operation of the group $G$? It does.

 

[xaveir]

Dummit and Foote clearly define it using the operation of that same group. What book are you using?

Are you suggesting that your book defines a group action of $(G,\ast_1)$ on $X$ via a function $\phi : G \times X \to X$ such that $\phi(g,\phi(h,x))=\phi(g\ast_2h,x)$, where $g,h\in G; x\in X,$ and where $\ast_2$ is the operation of another group defined on the elements of G? Because this definitely be a typo, as this would simply correspond to the usual definition of a group action of $(G,\ast_2)$ on $X$.

 

[Bachelier]

I tend to agree with you guys. It seems the operation is not really that important though as long as the action on the set is well-defined.

For the sake of discussion, I am including an image of the Author's (A.J Green) definition.