TITLE: Intuition behind Group Action
QUESTION [14 upvotes]: I am going over the group action and unfortunately I am not sure if I understand it very well. So for this reason, I have a few questions that possibly could help me to understand this concept better. A group action is defined as: Group $G$ acts on a set $X$ if there is a homomorphism $\sigma:G \rightarrow S_X$. I am aware group action has other definitions but for me this one is easier to understand.
Now $S_X$ is the group of permutations of $X$. It is not crystal clear what is meant by this to me. Is $S_X$ the same thing as $S_{|X|}$? When we talk about $S_n$, I have a clear mental picture; $n$ is an integer and I think of all the permutations with $n$ elements. But in $S_X$, $X$ is not an integer but a set. So suppose $X=\{1,2,3\}$. Is $S_X$ the same thing as $S_3$ in this case?
The other part of my question is about what the definition implies: it says "... if there is a homomorphism". Is it possible that homomorphism does not exist? Is the homomorphism unique? Or is it possible to have more than one homomorphism? Once we specify a group and a set, how do we find such homomorphism(s)? Is there a requirement for the size of the group and the set to make this definition work, for instance $|G| \geq |X|$?
Please let me know if you would like me to clarify my question. Any help with making a better intuition is appreciated!

REPLY [2 votes]: As for the first part of your question. If $X$ is finite of size $n\in\Bbb N$, then there is a bijection $f:\{1,\dots,n\}\to X$; build the map $\varphi\colon S_X\to S_n$ by $\varphi(\sigma):=f^{-1}\sigma f$; note that:

$f^{-1}\sigma f\in S_n$ (good definition);
$\varphi(\sigma)=\varphi(\tau)\Rightarrow f^{-1}\sigma f=f^{-1}\tau f\Rightarrow \sigma=\tau$ (injectivity);
for every $\alpha\in S_n$, $\alpha=\varphi({f\alpha f^{-1}})$ (surjectivity);
$\varphi(\sigma\tau)=f^{-1}\sigma\tau f=f^{-1}\sigma(ff^{-1})\tau f=(f^{-1}\sigma f)(f^{-1}\tau f)=\varphi(\sigma)\varphi(\tau)$ (group homomorphism)

Therefore, $S_X\cong S_n$ (noncanonically, being the isomorphism dependent on the bijection $f$).