Difference between the definition of monoid action and group action? The question is essentially in the title. From what I read in the wikipedia article about monoids it seems to me that we can define a monoid action in the exact same way we define a group action. Is that really the case or are there any pitfalls in this?
Thank you.
 A: Sure, you can do this.
Because the elements of a monoid don't necessarily have inverses, the map $f_g:X\to X$ is not necessarily $1-1$ or onto for any particular $g$.
Essentially, given a set $X$, we can see a "group action" as a homomorphism from $G$ to $\mathrm{Aut}(X)$, the group of $1-1$ and onto maps from $X$ to $X$. 
In the cases of monoids, instead of the group $\mathrm{Aut}(X)$, we use $X^X$, the set of all maps from $X$ to $X$, which is a monoid under function compositions.
Then given a monoid $M$, a monoid action on $X$ is a monoid homomorphism $\phi:M\to X^X$.
Indeed, you can find that every monoid is isomorphic to some sub-monoid of a monoid of the form $X^X$. In that sense, every associative operation with identity can be "faithfully represented" as function composition.
(Since every semigroup is easily a sub-semigroup of a monoid, you can actually make this a statement merely about associativity.)
A: A monoid action of a monoid $\mathcal M$ is a (left) semigroup action of $\mathcal M$ with the additional property that $\forall x\in X,e\cdot x=x$. A (left) semigroup action is the analogue, in semigroup theory, of a (left) group action. So, yes, you are right. I don't think that there are any pitfalls in this analogy.
