I have a feeling it will be tempting to mark my question as a duplicate, since I know this question is common (help to understand group actions), but my question is actually very specific. I have read about many different definitions, explanations, examples, and watched videos about group actions (including on this website) and had people explain it to me, and I feel like I understand what they are telling me but somehow I have some lingering misunderstanding. But, I finally figured out how exactly I want to ask my question, and it involves me explaining a little about what I do understand.
I am going to use the following definition of group action (but I know there are others which I have seen before); this one is from my textbook by Dummit and Foote:
$\underline{Definition}$ A group action of a group $G$ on a set $A$ is map from $G$ x $A$ (written as $g \cdot a$, for all $g \in G$ and $a \in A$) satisfying the following properties.
$1) \hspace{.2 cm} g_{1} \cdot (g_{2} \cdot a) = (g_{1}g_{2}) \cdot a$ for all $g_{1},g_{2} \in G, a \in A$, and
$2) \hspace{.2 cm} 1 \cdot a=a,$ for all $a \in A$
Now, my question involves the different notations between $g_{1} \cdot (g_{2} \cdot a)$ and $g_{1}(g_{2}a)$. I effectively want to know how equating these two expressions is different than proving associativity (and, i know that my last sentence is slightly different than part 1 of the definition of group action). My understanding is that the "dot" symbol ("\cdot" in latex notation) is supposed to mean "acting" and is sometimes written in different ways, but ultimately it could be a different operation than the operation defined as the group operation of $G$.
…I do understand that the set $A$ does not need to be a group or a subset of $G$, and so we can't define the $\cdot$ operation as being the same as the group operation of $G$. But, doesn't that make my above definition from Dummit and Foote incomplete or circular, since it doesn't define the operation $\cdot$? It seems like it's telling me to use the idea of "acting" in order to decide whether there is an action…which is circular, right?
…My ultimate question is how exactly part $1)$ of the above definition is different from group associativity, when $\cdot$ is not explicitly defined in most of the problems I have seen? It this above definition just so general, that it can't be used unless a specific example is given where $\cdot$ is defined?
I thought I understood this all before (group actions), and I recall that it helped me to understand that a group operation is an action from $G$ on itself. But my problem is figuring out the operation when $G$ is not acting on itself, when many examples and problems that I see ALSO do not define the operation $\cdot$. I would think you have to explicitly define $\cdot$ in order to even talk about whether something is a group action.
Also…as I was typing all that, I think I learned one more thing…a group action is actually a mapping, correct? It's not an element, or even a binary operation at all (I was thinking of it as a binary operation). And, a map is defined by some combinations of operations of elements (say, conjugation of an element by another element, or multiplication of an element by another element, or whatever any map might be defined as), and so, is this why the operation $\cdot$ that I keep talking about is never explicitly defined in a problem? Is it because $\cdot$ is referring to the relationships defined by the mapping, which would essentially be the thing I've been referring to as the "operation" $\cdot$?
Best Answer
Let’s forget about groups for a minute and talk about functions. Specifically we’ll consider some set $A$ and talk only about functions $f\,:\,A\to A.$
Hopefully this is starting to look a little familiar. We see this mix of two different things: one is composing two functions, an operation between two functions that gives us a new function; the other is applying functions to values, a strange “operation” of a function and some $x\in A$ that gives another element of $A$.
Now if we consider all the functions from $A$ to $A$ with inverses then we see they form a group under composition (hopefully it should be clear that composition must be associative). Let’s call this group $\mathrm{Sym}A.$
Now this is quite a nice group because we can somehow escape the group and have our group elements do something useful. But those group elements can do just about anything so it’s also useful to consider restricted subsets. For example if we take $A=\mathbb R$ we might only be interested in continuous functions, or if we take $A=\mathbb R^n$ we might only care about non singular linear maps (matrices), or if we take $A$ to be the vertices of a cube we might only be interested in functions that correspond to rotating the cube (so that each vertex lands in the place where another one used to be).
We see that these concepts would correspond nicely to other groups. The group action is a way of merging the two.
I don’t like the definition in terms of the weird $\cdot$ operation as a map $G\times A\to A.$ Morally, I think of it as “let’s secretly make every group element be a function and we’ll write function application either in the normal way with brackets or using a $\cdot$ symbol.” Now you can’t just go around making any group element into any old function willy nilly. You have to do it in a way that makes sense under the group laws. This should hopefully make the rules for a group action obvious. Clearly we must have $(g_1g_2)(x) = g_1(g_2(x))$ because the if the group elements secretly become functions then the group operation secretly becomes function composition.
Now that we’ve tried to give an intuitive idea of what a group action is, how can we formalise all this “every X lives a secret double life as a Y?” Well the answer is that we do it the same way we always have “some group somehow acts like some other group.” We have
I like this definition because it doesn’t live in a weird world of extra operations and rules but makes the point of group actions more clear and gives you all the rules for free from what we already know.
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