Semigroup partitions and Identity element

mnb96
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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?
 
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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.
 
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?
 
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.
 

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