TITLE: Does orbit-stabilizer theorem holds for monoid action?
QUESTION [7 upvotes]: For a group $G$ acting on some space $X$ we know there is a orbit-stabilizer theorem.
My question is does this formula holds for monoid action?
I think this formula may not hold, as inverse do not exist for all elements.
Though I am not sure.
My second question is what information do we get from monoid action? How it is useful?

REPLY [2 votes]: Monoid actions lead to the notion of a transformation monoid.
An interesting case is the following. A monoid $M$ acting on itself by right multiplication defines the (right) Cayley graph of $M$, having $M$ as set of vertices and edges of the form $(u, v, uv)$, where $u, v \in M$. The orbit of an element $m$ is the right ideal $mM$ and its (right) stabilizer is the set $\text{Stab}(m) = \{x \in M \mid mx = m\}$. The orbit-stabilizer theorem does not hold anymore, even if $M$ is finite and commutative. For instance, if 
$M = \{1, a, \dotsm, a^n \}$ with $a^{n+1} = a^n$, then for $m = a^{n-1}$, one gets $mM = \{a^{n-1}, a^n\}$ and $\text{Stab}(m) = \{1\}$, whence 
$|\text{Stab}(m)||mM| = 2$ although $|M| = n + 1$.
A systematic study of finite transformation monoids (and more generally of transformation semigroups) can be found in [1, 2]. There is also a large literature on the subject, notably in connection with algebraic automata theory.
For a more advanced treatment, see [3].
[1] S. Eilenberg, Automata, languages, and machines, Vol. B, Academic Press, New York, 1976.
[2] G. Lallement, Semigroups and Combinatorial Applications, Wiley and Sons, 1979.
[3] J. Rhodes, B. Steinberg, The $q$-theory of Finite Semigroups, Springer, Monographs in Mathematics, 2009