An extension of Lagrange's theorem to semigroups? The question is fairly dry: Is there any semigroup analogue of Lagrange's theorem for groups (counting as a generalization of the latter)? Let me guess the answer: Obviously yes. So the real question is: Any reference? Thank you in advance.
P.S.: I've notice of a Lagrange's theorem for Smarandache semigroups, but I would like to hear of different extensions, if possible (I don't think this is quite standard, but somebody defines a Smarandache semigroup to be any semigroup $(A, \star)$ for which there exist a proper subset $G$ of $A$, a unary operation $u: G \to G$ and a distinguished element $e \in G$ such that $(G, \star, u, e)$ is a group).
Edit. This is basically a comment to the subsequent answer of Vladimir Dotsenko. Let me highlight that I'm not asking for (possible) extensions to arbitrary semigroups. And I don't expect that, if any non-trivial extension is possible, it looks exactly like Lagrange's theorem for groups. 

I'm just asking for any possible
  non-trivial extension that is already
  there, in the literature. Say, for
  instance, an extension to some
  interesting classes of semigroups (apart from groups and those where the theorem sounds true by definition, e.g. Smarandache lagrangian semigroups).

I know, non-trivial and interesting are not well-defined terms. But I have faith in your common sense.
 A: There is an extension due to Schutzenberger. One needs to interpret Lagrange correctly. A partial function on a set $X$ is a function defined on a subset of $X$ to $X$. The composition of partial functions is defined where it makes sense. 
The set of all partial functions on X is a semigroup and so we can define the notion of an action of a semigroup $S$ by partial functions on the right of X. Say the action is transitive if for all $x\neq y$ in X there is $s$ in S with $xs=y$. 
Define an endomorphism of this action to be a totally defined map $f:X\to X$ such that $f(xs)=f(x)s$ (where equality means both sides are either undefined or agree). 


Theorem. If $X$ is finite and $T$ is a semigroup of endomorphisms of the action of $S$ on $X$, then the size of $X$ is divisible by the size of $T$. 


The proof is trivial: Show $T$ is a group acting freely on $X$. 
This generalizes Lagrange by taking $S=G=X$ with the regular right action and $T=H$ a subgroup acting on the left. 
Schutzenberger used this to generalize the monomial representation of groups to semigroups. 
A: I am a bit puzzled by your question. Do you mean the Lagrange's theorem stating that the order of a subgroup divides the order of the group? In that case, even for the semigroups defined in your second paragraph an analogue of Lagrange's theorem does not necessarily hold. Indeed, let's consider $\mathbb{Z}/6\mathbb{Z}$ as a semigroup with respect to the product. It has sub-semigroups of all possible orders: $\{0\}$, $\{0,1\}$, $\{0,1,3\}$, $\{0,1,2,4\}$, $\{0,1,2,3,4\}$, $\{0,1,2,3,4,5\}$ are all sub-semigroups, as one immediately checks. (And of course, invertible elements of $\mathbb{Z}/6\mathbb{Z}$ form a group, so your second paragraph cannot be literally true.)
A: Maybe this old paper will be of interest hor you:
Tamura, Takayuki
Note on finite semigroups which satisfy certain group-like conditions. (English)
Proc. Japan Acad. 36, 62-64 (1960).
