Using Dummit & Foote in a graduate level abstract algebra class, and it is my first introduction to group actions. I'm getting a better understanding as I go along, but one thing still puzzles me:
While I understand why care must be taken to differentiate between the inherent binary operation of the group and that of the group action (especially in terms of notation), for me it gets a little fuzzier when the group action is from a subgroup to the group in question. For example let $G$ be a group and $H\leq G$. Define the group action $\phi$:$H\times G\to G$ by $(h,g)\mapsto h\cdot g$; i.e., left-multiplication on $G$ (I'm pulling this scenario from Dummit & Foote, too – and here they even call $\cdot$ left-multiplication).
So, at this point with my example, can there really be any difference between $\cdot$, the operation of the group action, and that of $G$? Certainly $h\cdot g \in G$, and of course $hg\in G$ by closure (using concatenation to denote the binary operation of $G$). Since D&F called $\cdot$ left-multiplication, is it not true that $h\cdot g =hg$? What about in the general case, i.e. any group action from a subgroup to a group? Is the group action operation the same as the binary operation of the group?
Best Answer
They're different since $H\times G$ for $H<G$ is not $G\times G$.
Moreover, the use of $\cdot$ is a little like the definition
$$\begin{align} +:\Bbb R\times\Bbb R & \to \Bbb R,\\ (r,s) &\mapsto r+s \end{align}$$
compared to
$$\begin{align} +\rvert_{\Bbb Z}:\Bbb Z\times \Bbb R&\to \Bbb R,\\ (r,s)&\mapsto r+s. \end{align}$$