Group actions, permutation representations and currying I have been studying group actions for about a month, so I am at a basic stage and forgive me if this turns out to be a stupid question. It is about the two equivalent definitions of group action and permutation representation. In the 'action' definition of a left group action, we use a map $ \alpha: G \times X \rightarrow X $, which I understand can be curried to give a chain of two functions $ \phi : G \rightarrow (X \rightarrow X) $, which leads to the alternative definition of the group action as a permutation representation $ \phi : G \rightarrow Sym(X)$.
But how does this work for the right group action $ \alpha: X \times G \rightarrow X $, as following the same currying technique would give a chain of functions $ \phi: X \rightarrow (G \rightarrow X)$ and not a permutation representation?
In addition, what are the main reasons for wanting two equivalent definitions as an action and permutation representation? It seems to me that the definition as a group representation is quite natural, whereas the action definition is axiomatic and seems to come from nowhere. Many thanks.
 A: Those are not precisle equivalent definitions: any group action on a set determines a homomorphism from the group to the group of symmetries (permutations) of the set, and the other way around is also true.
Thus, if $\,\phi: G\times X\to X\,$ is a group action on a set $\,X\,$ (and it really doesn't matter whether this is a right or left action), then we get a homomorphism
$$\Phi:G\to\operatorname{Sym}_X\,\,\,,\Phi(g)(x):=\phi(g,x)$$
and this means that for any $\,g\in G\,$, the permutation $\,\Phi(g)\,$ is such that it maps and element $\,x\in X\,$ to the element $\,\phi(g,x)\,$ in $\,X\,$. 
It's easy to check the above is really a groups homomorphism, and also that we can do the same thing the other way around: begin with a homom. $\,G\to\operatorname{Sym}_X\,$ and get an action $\,G\times X\to X\,$ uniquely determined by the homomorphism.
A: Your observation is correct, indeed left group actions are exactly those functions corresponding to group homomorphisms (seen as functions from $G$ to the endofunction monoid $\text{End}(X)=\mathbf{Set}(X,X)$) via the following adjunction.
$$\mathbf{Set}(G \times X, X) \cong \mathbf{Set}(G,\text{End}(X))$$
This correspondence extend to right group actions, but the object which correspond to these actions are the antihomomorphisms of groups.
In general an antihomomorphism between the two groups $G$ and $H$ is just a function $f \colon G \to H$ such that for each pair $x,y \in G$ the equality
$$f(xy)=f(y)f(x)$$
holds.
Another way to state this is to see an antihomomorphism as a group homomorphism from the opposite group of the $G$, namely $G^\text{op}$ to $\text{Sym}(X)$.
$G^\text{op}$ is the group having as underlining set $G$ (so $G^\text{op}=G$ as set) having as multiplication the operation $\cdot \circ \sigma \colon G^\text{op} \times G^\text{op} \to G^\text{op}$ (where $\cdot \colon G \times G \to G$ is the group multiplication of $G$ and $\sigma \colon G \times G \to G \times G$ is the bijection sending the pair $(x,y)$ in the pair $(y,x)$, it's easy to prove that this is a group operation).
For every $r \colon X \times G \to X$ right action of $G$ on $X$ we can get the map $r \circ \sigma \colon G^\text{op} \times X = G  \times X \to X$ which is a left action of the group $G^\text{op}$ on the set $X$. So by what you told in your answer this action via currying correspond to homomorphism $G^\text{op} \to \text{Sym}(X)$ which are antihomorphisms from $G$ to $\text{Sym}(X)$.
Hope the answer was sufficiently clear and helpful.
