An analogy between group actions and group represenations I was trying to make a 'dictionary' between group action and group representation terms using the $\mathbb{C}[-]$ functor. I immediately found that if the set $Y \subset X$ is invariant under the action of $G$, then the corresponding representation is reducible. This of course means that irreducible represenations cannot be born out of actions that have non-trivial invariant subsets. But then again, the action of $\mathbb{Z}/2^2$ on itself by left shifts is transitive, but the corresponding representation is of course reducible. More generally, if $G$ is abelian then it appears that no irreducible representation can come out of group action.
My question is: does it make sense to try and draw parallels further? I find it compelling that the functor takes $\sqcup$ to $\oplus$ and $\times$ to $\otimes$ (and these operations turn out to be important in representation theory), but the lack of an analog for irreducibility means that most representation theory will not correspond to anything.
 A: When you turn a finite $G$-set $X$ into a permutation representation $V=\mathbb{C}[X]$, then $V$ is never irreducible. Indeed, it always has the trivial representation as a summand, generated by $\sum_{x\in X} s$. You can still ask yourself what happens if you take this one summand out. The beautiful answer is that the remaining representation will be irreducible if and only if $G$ acts on $X$ doubly transitively. See Proposition 5.16 and the following corollary in these lecture notes on representation theory.
The precise relationship between $G$-sets and permutation representations is an active area of research. For example you can ask when two non-isomorphic sets give rise to isomorphic representations. This question is important not only in algebra, but also has applications to number theory, differential geometry, topology, and probably other areas. It took more than fifty years to answer this question in full generality, and the answer is very recent.
Also, if you replace $\mathbb{C}$ by $\mathbb{Q}$ and ask yourself, which characters of rational representations can be expressed as linear combinations of permutation characters, then you also quickly arrive at uncharted territory. See this question on math.se.
