Semigroup partitions and Identity element

[mnb96]

If I have a semigroup S, is it possible to partition the set of element S into two semigroups $$S_1$$ and $$S_2$$ (with $$S_1 \cap S_2 = 0$$), in such a way that $$S_1$$ has an identity element but $$S_2$$ has none?

 

[HallsofIvy]

If the orginal semigroup has an identity, yes! If your semigroup has an identity, but no element has an inverse, then taking S1 to be the identity only, S2 all other elements, works.

 

[mnb96]

I should have been more specific:

- The semigroup S has no identity (it just satisfies associativity and closure)
- I want to partition S into $$S_1$$ and $$S_2$$ such that:
- $$S_1$$ is a subsemigroup and has a right-identity for itself
- $$S_2$$ is a subsemigroup but does not have any identity element

Is this situation possible at all?

 

[mnb96]

If I worked out my proof correctly, then under those conditions, if $$S_1$$ had a right-identity, then $$S_2$$ must have at least a left-identity.

I based my proof on the observation: $$ab = (a1)b = a(1b)$$, with $$a\in S_1$$, $$b\in S_2$$

I'd like to be confirmed to be right anyways.