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.
Postscripts:
- 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 $f\cdot g\cdot h\cdot x$
- Each of the “subgroups of the group of functions” above is really a natural group action in disguise
- 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.
- Some properties of group actions come from properties of the homomorphism $\phi,$ for example an action is faithful if $\phi$ is an injection
- 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.
- 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.
If a group $G$ acts on another group $H$ (by automorphisms), then a distinguishing condition takes place, namely:
$$\operatorname{Fix}(g)\le H, \forall g\in G \tag 1$$
where $\operatorname{Fix}(g):=\{h\in H\mid \phi_g(h)=h\}$, being $\phi\colon G\longrightarrow \operatorname{Aut}(H)$ the action. In fact, for every $g\in G$, $\phi_g(1_H)=1_H$, whence $1_H\in\operatorname{Fix}(g)\ne\emptyset$; moreover, by definition, $\operatorname{Fix}(g)\subseteq H$; finally, for every $g\in G$ and $h_1,h_2\in\operatorname{Fix}(g)$, $\phi_g(h_1h_2^{-1})=$ $\phi_g(h_1)\phi_g(h_2^{-1})=$ $\phi_g(h_1)\phi_g(h_2)^{-1}=$ $h_1h_2^{-1}$, whence $h_1h_2^{-1}\in\operatorname{Fix}(g)$.
If $G$ and $H$ are both finite, say $G=\{g_1,\dots,g_{|G|}\}$ and $H=\{h_1,\dots,h_{|H|}\}$, then the usual equation:
$$\sum_{i=1}^{|H|}|\operatorname{Stab}(h_i)|=\sum_{j=1}^{|G|}|\operatorname{Fix}(g_j)| \tag 2$$
and the usual condition (by Burnside's Lemma):
$$|G|\mid \sum_{j=1}^{|G|}|\operatorname{Fix}(g_j)| \tag 3$$
give new opportunities, according to the fact that -by Lagrange's Theorem- the summands in the RHS of $(2)$ and in $(3)$ must all divide $|H|$.
As an entry test for this new setting, let's take $G=C_p$ and $H=C_q$, where $p$ and $q$ are distinct primes. If a nontrivial homomorphism exists, then $|\operatorname{Fix}(g_{\bar j})|=1$ and $|\operatorname{Stab}(h_{\bar i})|=1$, for some $\bar j\in \{1,\dots,p\}$, $\bar i\in\{1,\dots,q\}$. Then $(2)$ and $(3)$ yield:
$$[\space k+(q-k)p=l+(p-l)q\Longrightarrow k(p-1)=l(q-1)\space]\space\wedge\space [\space p\mid pq-l(q-1)\space ] \tag 4$$
for some $1\le k\le q$ and $1\le l\le p$. Now:
- if $p>q$, then from $(4)$-2nd term of the "$\wedge$": $p\mid pq-l(q-1)\Longrightarrow$ $p\mid l \Longrightarrow$ $l=p\Longrightarrow$ $k(p-1)=p(q-1)\Longrightarrow$ $k=\frac{p}{p-1}(q-1)>q-1\Longrightarrow$ $k=q$; but $(k,l)=(q,p)$ is not a solution of $(4)$-1st term of the "$\wedge$": so, for $p>q$, there are no nontrivial homomorphisms $\phi\colon C_p\longrightarrow\operatorname{Aut}(C_q)$;
- if $p<q$ and $p\nmid q-1$, then from $(4)$-2nd term of the "$\wedge$": $p\mid pq-l(q-1)\Longrightarrow$ $p\mid l$, and we fall back into the previous case.
Therefore, if $p\nmid q-1$, there are no nontrivial homomorphisms $\phi\colon C_p\longrightarrow\operatorname{Aut}(C_q)$. This hasn't used any knowledge about the isomorphism class of the group $\operatorname{Aut}(C_q)$. (Incidentally, that for $p\mid q-1$ there is actually a nontrivial homomorphism $\phi$, it is shown e.g. here.)
As one more application, let's now assume $G=C_p$ and $|H|=q^2$, still with $p,q$ distinct primes. If a nontrivial homomorphism exists, then $(2)$ and $(3)$ yield:
$$[\space k+(q^2-k)p=l_0+l_1q+(p-l_0-l_1)q^2\space]\space\wedge\space [\space p\mid k\space ] \tag 5$$
for some $1\le k\le q^2$ and $1\le l_0+l_1\le p$. Since $p\mid k\Longrightarrow p\le k\le q^2$, for $p>q^2$ there are no nontrivial homomorphisms $\phi$. This suffices to give a different proof to, e.g., this question (and the conclusion doesn't change with $\Bbb Z_2\times \Bbb Z_2$ in place of $\Bbb Z_4$). Again, this hasn't used any knowledge about the isomorphism class of the group $\operatorname{Aut}(H)$.
Best Answer
All the orbits are the conjugacy classes of an element in $G$. You can divide elements in $G$ into two parts, those that are in the center $Z(G)$ of $G$, and those that are not in the center.
Let $x \in Z(G)$, then $x$ is fixed by conjugation by any elements in $G$. Hence when you look at the orbit-stabilizer formular, you have that $|G|=|Z(G)|+\sum_{x \in conjugacyclasses}{\frac{|G|}{|stab(x)|}}$. you can use this formula to show for example group of order $p^2$ is abelian as $p||G|$ and $p|\frac{|G|}{|stab(x)|}$ for all $x$ in the sum. so $p||Z(G)|$. Hence $G$ has no trivial center. And you see that $G \cong Z/p^2Z$ or $G \cong (Z/pZ)^2$ which are both abelian.