This question already has an answer here:

This question is not a duplicated of https://math.stackexchange.com/a/3183330/342834 . It is a follow-on, asking about the distinction between the subject of that post, and the subject of this post.

In The Number System by Thurston the author introduces an algebraic structure he calls a hemigroup. The laws of a hemigroup are:

  • (i) $\left(x\odot y\right)\odot z=x\odot \left(y\odot z\right),$
  • (ii) $\left(x\odot y\right)=\left(y\odot x\right),$
  • (iii) $\left(x\odot y\right)=\left(x\odot z\right)\implies{y=z},$
  • (iv) $\exists_e e\odot e=e.$

I have been told this defines a cancellative commutative monoid. I will retain the uncommon designation hemigroup for now.

Thurston then introduces entities he calls dyads which are sets of ordered pairs formed of elements of the hemigroup constituting equivalence classes of the ordered pairs. I write this as

$$\left[\![a,b\right]\!]\equiv\{ \left<x,y\right>\backepsilon{x\odot b=y\odot a}\}.$$

The algebra of these dyads satisfies the laws of a commtative group. One such commutative group is what BBFSK call the module of integers. In that source, however, this module is constructed of residue classes of ordered pairs from an algebraic structure which amounts to Thurston's hemigroup without the existence of an identity element. That is

  • (i) $\left(x\odot y\right)\odot z=x\odot \left(y\odot z\right),$
  • (ii) $\left(x\odot y\right)=\left(y\odot x\right),$
  • (iii) $\left(x\odot y\right)=\left(x\odot z\right)\implies{y=z}.$

But those authors do not formally nominate this structure. It is the additive algebraic structure of the natural numbers, which by traditional American definition are $\mathbb{N}\equiv\{1,2,3,\dots\}$. Thurston's hemigroup is the additive structure of the whole numbers, which by traditional American definition are $\mathbb{N}_0\equiv \mathbb{N}\cup\{0\}.$

Both developments produce the same integral domain, but from different underlying algebraic structures.

I am asking if the algebraic structure consisting of a hemigroup sans identity has a common mathematical name. If so, what is that name?


marked as duplicate by YCor, Lord Shark the Unknown abstract-algebra Apr 25 at 2:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 3
    $\begingroup$ It's called a commutative cancellable semigroup. By the way, axiom (iv) as stated, doesn't imply the existence of an overall identity, it only requests an idempotent element to exist. $\endgroup$ – Berci Apr 24 at 16:02
  • $\begingroup$ Thank you for that clarification regarding the idempotent element. Thurston, in fact, never uses that element directly, but only in defining the isomorphism between dyads of the form $\left[\![x,e\right]\!]$ and the underlying hemigroup, as well as defining the identity dyad $\varepsilon\equiv \left[\![e,e\right]\!]$. For the latter $e$ is certainly unnecessary. $\endgroup$ – Steven Thomas Hatton Apr 24 at 16:15
  • 2
    $\begingroup$ Looks like almost a duplicate of your recent previous question Is there a more common term for "hemigroup". Please rather edit your previous question. $\endgroup$ – YCor Apr 24 at 16:45
  • $\begingroup$ These are distinct questions. The first asked for a commonly used term for Thurston's hemigroup. This question is asking for a term for a structure which is similar to, but distinct from a hemigroup. $\endgroup$ – Steven Thomas Hatton Apr 24 at 17:26