Need help understanding Group Actions

Question

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" $\cdot$?

Answer 1

Let’s forget about groups for a minute and talk about functions. Specifically we’ll consider some set $A$ and talk only about functions $f\,:\,A\to A.$

• we have an identity function $1(x) := x$
• we can compose two functions $(f\circ g)(x) = f(g(x))$
• sometimes a function $f$ has an inverse $f^{-1}$ where $f(f^{-1}(x))=x=f^{-1}(f(x)),$ i.e. $f\circ f^{-1}=1=f^{-1}\circ f.$

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

Definition: A group action of a group $G$ on a set $X$ is a group homomorphism $\varphi :G\to\mathrm{Sym} A.$ We write $g\cdot x$ or $g(x)$ or when we are more comfortable $gx$ for $\varphi(g)(x).$

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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Postscripts:

1. I think to answer your question about the lack of any associativity rule for group actions I should just point out that $\cdot:G\times A\to A$ doesn’t have enough symmetry to warrant an extra rule. There is only one way to interpret $f\cdot g\cdot h\cdot x$
2. Each of the “subgroups of the group of functions” above is really a natural group action in disguise
3. It turns out that as well as there being lots of natural actions of “groups on things” it is often useful to consider actions of groups on groups or group-related things like cosets.
4. Some properties of group actions come from properties of the homomorphism $\phi,$ for example an action is faithful if $\phi$ is an injection
5. There is also a concept called a right action which is a bit more fiddly to get your head round if you think of actions as turning group elements into functions (with application on the left). It’s basically like $x\cdot g=\varphi(g^{-1})(x).$ Such actions come up occasionally. I like to imagine pushing your elements of $A$ through the group elements in the action direction (left or right), flattening and removing the group elements as we push through.
6. A generalisation of action replaces $\mathrm{Sym}A$ with other general transformation groups. For example a group representation is like a “linear action”. It is a group homomorphism into a general linear (ie matrix) group.

Answer 2

Associativity of the group requires that, for $g_{1}, g_{2}, g_{3} \in G$, we have $(g_{1}g_{2})g_{3} = g_{1}(g_{2}g_{3})$.

For the above property of a group action, the requirement is different because the element $a$ is not thought of as being in $G$. We instead have an "action" of the elements of $G$ on $A$, in other words think of each element of $G$ as a bijection from $A \to A$. So the notation is more requiring that $g_{1}(g_{2}(a)) = (g_{1}\cdot g_{2})(a)$, where the $\cdot$ represents group multiplication.

The notation is a little awkward, because it is common to use $g_{1}g_{2}$ and $g_{1} \cdot g_{2}$ interchangably for group multiplication; however here they are using $\cdot$ for the action of a group element $g$ on an element $a \in A$ (as in $g \cdot a$), and not between group elements.

Answer 3

I like to think about a group action in the following way. For each $g\in G$, we obtain a permutation of $A$ via $\sigma_{g}(a):=g\cdot a$ for all $a\in A.$ In order for this map ($g\mapsto \sigma_{g}$) to be a group action, we require that 1) $\sigma_{g'}(\sigma_{g}(a))=\sigma_{g'g}(a)$ for all $a\in A,$ and for any $g,g'\in G,$ and 2) $\sigma_{e}(a)=a$ for all $a\in A,$ where $e$ is the identity in $G$. In other words, if we let $X$ be the group of permutations of the elements of $A$, then this equality says that $g\mapsto \sigma_{g}$ must be a group homomorphism $G\rightarrow X.$

Answer 4

Let $m$ be a map from $G\times A$ to $A$. Your quoted definition doesn’t point out, which is unfortunate, that we traditionally write $m((g,a))$ (the image of $(g,a)\in G\times A$ under $m$ — and which is an element of $A$) as $g\cdot a$. The map $m$ is a group action if $m((g_2,m((g_1,a))))$ is always $m((g_2g_1,a))$ and $m((1,a))$ is always $a$. Using the $\cdot$ notation simplifies writing the definition, but it reveals no name or notation for the actual group action map from the Cartesian product $G\times A$ to the set $A$. I think that might be the reason for your trouble understanding this.