TITLE: What's the idea of an action of a group?
QUESTION [13 upvotes]: I know the formal definition of an action over a set. I'm not asking this. 
What I'm asking is: what's the intuition of it? It is a way to define an algebra over a set? Since an action can exist in the most arbitrary of sets, does this mean that every set can be assumed as with an algebraic structure? Is it not possible for a set to be "unactionable" due to its structure? For example, if I have a group G with elements {a,b,c...} and a set M, and we let G act in M, how can there be such an element $a\cdot x$ in M if $a$ and $x$ are of different natures (like $a$ being a matrix and $x$ a real number)? Do I have to find a group $G'$ which is isomorphic to $G$ in such a way that this action makes sense over M? What if this isomorphism is unobtainable?

REPLY [3 votes]: Let's say we have a set of interesting things; for example, we might have the set $C$ of faces of a cube, or the set $\mathbb{R}^3$ of points in Euclidean 3-space.
To any set $S$, there is a group denoted $\mathsf{Aut}(S)$ which consists of all bijective functions $f:S\to S$, where the operation is $\circ$ (composition), and the identity is just the identity function $\mathrm{id}_S(s)=s$.
However, it's usually not the case that we care about every possible way of rearranging (i.e., permuting) the elements of our set; for example, the permutation of $C$ swapping two adjacent sides and doing nothing else is not really as natural a thing to consider as the permutation of $C$ that a rotation of the cube induces. And there are tons of crazy ways of permuting $\mathbb{R}^3$ that can't even be written down; but we're mostly interested in mapping $\mathbb{R}^3$ to itself in geometrically interesting ways, like reflection or rotation.
Thus, we decide to restrict our attention to a certain subgroup $G$ of $\mathsf{Aut}(S)$.
That is (in essence) a group action: we have a group $G$, each of whose elements determines a permutation of $S$.
In the case of $\mathbb{R}^3$, we might decide in advance that the only permutations we are interested in are bijective linear transformations from $\mathbb{R}^3$ to itself, since after all $\mathbb{R}^3$ has this nice vector space structure, and we might as well preserve it. Then we are doing representation theory; we have a group $G$, each of whose elements is assigned a linear isomorphism from $\mathbb{R}^3$ to itself, which (thought of as a matrix) is an element of $\mathsf{GL}_3(\mathbb{R})$.
Now, sometimes we start out with the group $G$ already in mind. Maybe it came up in a problem we are doing, and we noticed that there is a natural way that its elements can give rise to permutations of some set (or linear isomorphisms of some vector space). Then, what we have is not a subgroup $G\subset \mathsf{Aut}(S)$ or $G\subset\mathsf{GL}_n(\mathbb{R})$, but instead, a homomorphism $\rho:G\to\mathsf{Aut}(S)$ or $\rho:G\to\mathsf{GL}_n(\mathbb{R})$.
Note that there is a phenomenon that can occur in this more general case, that does not occur in the way I initially described things, that deserves to be mentioned: multiple group elements can correspond to the same permutation (or linear isomorphism). In other words, the homomorphism $\rho:G\to\mathsf{Aut}(S)$ might have a non-trivial kernel.