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First let $(G, \circ)$ be a group and let $S = G$. Then let $g,h \in G$ and $s \in S$.

I'm confused by the meaning of the statement "$G$ acts on itself by right multiplication". (Specifically, I'm being asked to show "G acts on itself by right multiplication" without any context). It seems to me this could mean two things:

(1) The function $G \times S \rightarrow S$ s.t. $gs = s \circ g^{-1}$ forms a left group action on G.

(2) The function $S \times G \rightarrow S$ s.t. $sg = s \circ g$ forms a right group action on G.

Is there a standard meaning to this question or should I ask my instructor?

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    I think either of these interpretations could be reasonable. Some people use the term action as a shorthand for left action, in which case they would mean (1). On the other hand, because (2) doesn't require the use of the inverse in the definition, this could be considered the 'simplest' interpretation of the statement. @OP: You should have some context available, namely what the definition of the word 'action' is in your course. If you are not sure, asking the instructor seems reasonable.2012-12-17

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You describe the case that the group $G$ acts (left/right) on the set $S$. Here we are concerend with the case that $S$ is the group $G$ itself (or more precisely, the underlying set of the group). Thus we are letting $sg=s\cdot g$ (i.e. we multiply at the right) for $s\in S=G$ and $g\in G$. You need to show that the axioms of action are fulfilled by this.

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    That's a good example why writing just the underlying set for the whole group can be quite misleading for people new to group theory.2012-12-17