I'm just starting to learn about group actions and there is something not clear to me.
Let $S = \{1, \dots, n\}$ and let $G$ be a group acting on $S$.
Does this imply that $G$ is a permutation group, that is, that $G \subseteq S_n$ where $S_n$ is the symmetric group? Or could $G$ be any group like e.g. the Klein 4 group?
Given any set $S$, any group $G$ can act on $S$ via the trivial action ($gs=s$ for all $g\in G$ and $s\in S$). Thus, any group can act on $\{1,\ldots,n\}$, regardless of whether it is an actual, literal subset of $S_n$ (the group of permutations of the set $\{1,\ldots,n\}$).
On the other hand, every group does embed in (i.e., have an injective homomorphism to) some group of permutations, namely, the group of permutations of its underlying set: $$\varphi:G\longrightarrow S_G,\qquad \varphi(g)=\text{ the permutation }\sigma:G\to G \text{ defined by }\sigma(h)=gh$$ Thus, for example, if $G$ is a group with $n$ elements, then we can naturally view $G$ as a subgroup of $S_n$. This result is usually called Cayley's theorem (Wikipedia link).