I’m taking a topology class and the professor decides to introduce a bit of group theory. We’re getting to group action and I find myself getting a bit confused about the language used in the definition. I will try to rephrase an example I found on wikipedia, and would like to ask for correction if necessary.
The example I’m referring to is the following picture in this link:
Now the introduction of the wikipedia article says that:
In mathematics, a group action on a space is a group homomorphism of a given group into the group of transformations of the space.
I assume in this example, the given group is $C_3$ (with group operation I denote $\cdot$), the space is the set of three vertices of the triangle, and the group of transformations of the space is $S_3$ (with group operation I denote $\star$). Then the group action would be a group homomorphism $\alpha: C_3 \times S_3 \to S_3$ that satisfies the identity and compatibility axioms of group action (here I follow the actual definition of group action given in the article).
There’s an intuitive understanding of what $\alpha(\rho, s)$ would do, for $\rho \in C_3$ and $s \in S_3$. However, I find it hard to verify that $\alpha$ is indeed a group homomorphism, as there’s no “formulae” for the group operations $\cdot$ and $\star$. Specifically, given $\rho$ and $\rho’$ in $C_3$, $s$ and $s’$ in $S_3$, and let $\ast$ be the group operation of $C_3 \times S_3$, I want to check that $\alpha((\rho,s) \ast (\rho’,s’)) = \alpha(\rho,s) \star \alpha(\rho’,s’)$. Here I find myself unable to proceed (except checking by hand), since I don’t know how to express the way any of the operations $\cdot$, $\star$ or $\ast$ work (except for $\ast$, where I think of doing thing component-wise, i.e. $(\rho,s) \ast (\rho’,s’)$ would be equal to $(\rho \cdot \rho’, s \star s’)$).
Also, is saying: “The cyclic group $C_3$ consisting of the rotations by $0^o$, $120^o$ and $240^o$ acts on the set of the three vertices” not technically correct? “The set of the three vertices” is not a group of transformations, and, itself not being a group, it doesn’t have a group operation with which I can check that the action is indeed a homomorphism?
Any idea how I can understand the above?
An action is a way to "using" the elements of a group (the "actors") to "move" each element of a set (the "acted upon") to one element of the same set. Therefore, an action is firstly and foremost a map $\mathcal{A}\colon G\times X \to X$. The usual pair of conditions qualifying such a map as an action are nothing else that the generalization to abstract groups of the basic properties ensured by the natural action of the "prototypical" group $\mathrm{Sym}(X)$, namely:
You can prove that, given an action $\mathcal{A}$ as above, the position $\varphi(g)(x):=\mathcal{A}(g,x)$ defines a group homomorphism $\varphi\colon G\to \mathrm{Sym}(X)$. Conversely, given a group homomorphism $\varphi\colon G\to \mathrm{Sym}(X)$, the position $\mathcal{A}(g,x):=\varphi(g)(x)$ defines a $G$-action on $X$.