Is a finite semigroup isomorphic to subsets of some group?

[Stephen Tashi]

Is any given finite semigroup isomorphic to some finite semigroup S that consists of some subsets of some finite group G under the operation of set multiplication defined in the usual way? (i.e. the product of two subsets A,B of G is the set consisting of all (and only) those elements of G that result from multiplying some element in A times some element in B using the group operation defined on G).

 

[micromass]

A necessary condition is of course that the semigroup is cancellative, since groups have this property: http://en.wikipedia.org/wiki/Cancellative_semigroup So this means that $ax = bx$ implies $a=b$ and $xa = xb$ also implies $a=b$.

If the semigroup is abelian, then the Grothendieck construction gives an explicit group where-in the semigroup is embedded. So for abelian semigroups, this is an equivalence.

Now, it turns out that any finite cancellative semigroup is already a group. So the answer to your question is that any finite semigroup that can be embedded in a group, is already a group. The latter is easily seen directly: let $G$ be a group and let $S$ be an embedded semigroup. Take an element $a\in S$, then

$$\{a,a^2,a^3,a^4,...\}\subseteq S$$

Since $S$ is finite, there exists some $n\neq m$ such that $a^n = a^m$. Say that $n>m$, then $e = a^{n-m}\in S$. Furthermore, $a^{n-m-1}$ is also in $S$ and this is an inverse.

 

[Stephen Tashi]

That was a very useful reply - and quick too!

I'm not clear yet on what an "embedding" must be and whether my question is more general than asking for an embedding.

A necessary condition is of course that the semigroup is cancellative, since groups have this property: http://en.wikipedia.org/wiki/Cancellative_semigroup So this means that $ax = bx$ implies $a=b$ and $xa = xb$ also implies $a=b$.

I agree that groups have the cancellative property, but does a semigroup whose elements are subsets of group (rather than being necessarily single elements of the group) have the property? For example, the usual definition for multiplying sets of group elements would make $\emptyset$ behave like a zero element. The presence of a zero would screw up cancellation.

 

[micromass]

So you have a group $G$, and you take $\mathcal{P}(G)$ and you put a product on there that says $AB = \{ab~\vert~a\in A,~b\in B\}$? Then no, this does not have the cancellative property since $GG = G = \{e\} G$.

I don't know the answer to your question. But it is important to note that the subsets of some group have a partial ordering relation given by $\subseteq$, such that if $H\subseteq H^\prime$, then $HK\subseteq H^\prime K$. So the order agrees with the multiplication operation.

So any semigroup that is isomorphic to the subsets of some group would have to admit such an order relation.

For abelian groups we can even say more and we can add that any two elements under the order relation must have a supremum.

 

[micromass]

Well, as far as I can see, the problem is quite difficult. Here is some literature on the subject:

http://link.springer.com/content/pdf/10.1007/BF02572473.pdf
http://dml.cz/bitstream/handle/10338.dmlcz/102069/CzechMathJ_36-1986-1_12.pdf