1
$\begingroup$

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.

Improve this question
$\endgroup$
1

3 Answers 3

Reset to default
6
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thank you, Benjamin. But I'm not quite sure to understand what you mean by saying that the composition of partial functions (from $X$ to itself) could make no sense in some cases. Partial functions (between sets) should be composed with each other in the same way as binary relations. Then, the corresponding category would be equivalent to the usual category of pointed sets. $\endgroup$
    – Salvo Tringali
    Commented Jun 15, 2012 at 20:08
  • $\begingroup$ I mean composition as binary relations. $\endgroup$
    – Benjamin Steinberg
    Commented Jun 15, 2012 at 21:07
  • $\begingroup$ Then, as for your case, the composition of two partial functions (from $X$ to $X$) makes always sense. Thanks for the clarification $\endgroup$
    – Salvo Tringali
    Commented Jun 15, 2012 at 21:47
  • $\begingroup$ I meant the composition of partial functions is defined on those points of X for which it makese sense. $\endgroup$
    – Benjamin Steinberg
    Commented Jun 15, 2012 at 21:49
1
$\begingroup$

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.)

Improve this answer
$\endgroup$
7
  • $\begingroup$ Yes, I mean that theorem. But Lagrange's theorem for Smarandache semigroups is not exactly what you're thinking of. In particular, it is not a statement about all the subsemigroups of a given semigroup. Indeed, it is not even about arbitrary Smarandache semigroups. Instead, it concerns a special class of them, which somebody calls Smarandache lagrangian semigroups. Thus, what they refer to as the Lagrange's theorem for Smarachande semigroups is essentially true by definition. And yes, I agree with you that this may be a -cypa- thing, but I'm not here to quibble in the value of others' work. $\endgroup$
    – Salvo Tringali
    Commented Jun 15, 2012 at 12:31
  • $\begingroup$ OK, what is your motivation for this question then? I think this example shows that there is no serious general result to hope for. If you are saying that in the only somewhat general result the theorem holds by definition, what makes you hope that something meaningful does exist? $\endgroup$
    – Vladimir Dotsenko
    Commented Jun 15, 2012 at 12:36
  • 1
    $\begingroup$ That's a good example. It's already clear from considering finite monogenic semigroups that the order of a subsemigroup of a 'Smarandache Semigroup' doesn't have to divide the order of the semigroup, but this example shows that nothing I would call an 'analogue of Lagrange's theorem' holds. $\endgroup$
    – Tara Brough
    Commented Jun 15, 2012 at 12:41
  • $\begingroup$ As for the rest, 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 the -cypa- ones where the theorem sounds true by definition. I know, non-trivial and interesting are not well-defined terms. But let me believe in your common sense. $\endgroup$
    – Salvo Tringali
    Commented Jun 15, 2012 at 12:43
  • $\begingroup$ (I was typing that comment quite slowly, so didn't see the two preceding comments until afterwards.) $\endgroup$
    – Tara Brough
    Commented Jun 15, 2012 at 12:44
1
$\begingroup$

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).

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .