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

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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.
 
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Bachelier said:
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.
 
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.
 
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.
 

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