Intuition on group actions I'm trying to get more intuition on this definition:   
Let $(G,{}\circ{})$ be a group with identity element $e$. 
A group action is a mapping $\varphi : G \times X \to X \,$ such that:
\begin{alignat}
{}& \forall x \in X, &\varphi(e, x) = x. \tag{1}\\
{}& \forall g,h \in G, \forall x \in X, &\varphi((g\circ h),x) = \varphi(g,\varphi(h,x)). \tag{2}
\end{alignat}
Is there some kind of intuition behind this definitions ? From the examples I've seen it seems like you always get a bijective mapping $X →X:x↦g(x)$. Is this the idea ? That $G$ acts on $X$ if all elements of $g$ are bijective mappings from $X→X$ ?
 A: Maybe I'm oversimplifying the matter, but the definition of action, which is property-based and not given by a closed formula, is patterned upon the basic properties fulfilled by the "prototypical" action, namely the natural acting (indeed) of the bijections on a set $X$ on the elements $x \in X$:
\begin{alignat}{1}
&\iota_X(x)=x, \forall x \in X \\
&(\sigma\tau)(x)=\sigma(\tau(x)), \forall \sigma,\tau \in \operatorname{Sym}(X), \forall x\in X
\end{alignat}
The generalization to abstract groups, represented by $(1)$ and $(2)$ in the OP, works because -in turn- the definition of abstract group is precisely patterned upon the properties fulfilled by a set of bijection endowed with the composition as operation (closure, associativity, identity, inverses).
A: Let $\mathcal{F}$ be the collection of functions that map $X$ to itself. There is a group in $\mathcal{F}$, that is, the collection of invertible functions, $\mathcal{G}$ (The unit is the identity function, the binary operation is composition). 
Now given a group $G$, we want to know whether there is a group homomorphism \begin{equation}
G\xrightarrow{h}\mathcal{G}. 
\end{equation} If there is, then the image is a subgroup in $\mathcal{G}$ that behaves like $G$, so probably we can learn something about $\mathcal{G}$ or $X$ from $G$. 
But now we have $h(e)$ is the identity function on $X$. Thus $h(e)(x)=x$ for all $x\in X$. Also, if $g_i\in G$, then $h(g_1g_2)(x)=h(g_1)\circ h(g_2)(x)$. These follow from the fact that $h$ is a homomorphism.
In particular, this $h$ defines a group action of $G$ on $X$. The other direction is just as natural.
So to conclude, by a group action we are just trying to find a subgroup of functions on $X$ that behaves like some group we already knew. Also note that the group of invertible functions are symmetries of $X$, we are looking at subgroups of symmetries of $X$, no wonder we can learn a lot about $X$ from the group actions.
In practice, we seldom just take $X$ as a set and look at all the invertible functions. We often impose more structures on $X$, and only look at the collection of functions that respect these structures. But the spirit remains.
For more examples and a much clearer exposition, look at this wonderful article by Gowers. 
