Semigroup partitions and Identity element

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Discussion Overview

The discussion revolves around the possibility of partitioning a semigroup into two subsemigroups, where one subsemigroup has an identity element and the other does not. The context includes theoretical considerations regarding semigroups, identity elements, and their properties.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant questions whether it is possible to partition a semigroup S into two subsemigroups S_1 and S_2, with S_1 having an identity element and S_2 having none.
  • Another participant suggests that if the original semigroup has an identity, a specific partition can be made where S_1 consists of the identity element and S_2 contains all other elements.
  • A later post clarifies that the original semigroup does not have an identity and asks if it is possible to create a partition where S_1 has a right-identity and S_2 has no identity element.
  • One participant claims that if S_1 has a right-identity, then S_2 must have at least a left-identity, based on their proof involving specific element interactions.

Areas of Agreement / Disagreement

Participants express differing views on the conditions under which such a partition is possible, with no consensus reached on the validity of the claims made regarding identities in the subsemigroups.

Contextual Notes

The discussion includes assumptions about the properties of semigroups, particularly regarding identity elements and their implications for partitioning. The proof mentioned relies on specific observations that may not be universally applicable without further clarification.

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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