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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 (see comments) that $e$ of (iv) is not necessarily an identity element, but merely an idempotent element. I'm trying to understand what this actually means. I assume the distinction would be that $x\odot e=e\odot x=e\odot x\odot e,$ etc., but we have no rule $x\odot e=x.$ Please correct me if this is incorrect.

My main question is this: if $\mathcal{H}$ is the set of the hemigroup and we define our set of interest to be $\mathcal{S}\equiv \{\xi\backepsilon \xi=x\odot e\land e,x\in \mathcal{H}\},$ preserving the meaning of $\odot$ in $\mathcal{S},$ does $e$ now constitute an identity element of $\mathcal{S}?$ I am accepting the identity $e=e\odot e\in\mathcal{S}$.

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  • $\begingroup$ The definition of algebraic structure given in BBFSK regarding sets of abstract mathematical entities is: "These sets have an algebraic structure consisting in certain relations or laws of combination among the elements within the set, in the existence of certain distinguished subsets, and so forth." In their formal exposition of structure the existence of an identity element is part of the specification of algebraic structure. So, by my book, there is no distinction. $\endgroup$ – Steven Thomas Hatton Apr 24 at 17:42
  • $\begingroup$ Can you explain the quantifier on x in axiom iii? $\endgroup$ – Tim Carson Apr 24 at 21:52
  • $\begingroup$ The axioms are for all $x,y,z\in \mathcal{H}$. $\endgroup$ – Steven Thomas Hatton Apr 24 at 23:36
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Actually, $e$ is an identity element of $\mathcal{H}$ itself, because of axiom (iii). For any $x\in\mathcal{H}$, $$e\odot x=(e\odot e)\odot x=e\odot(e\odot x)$$ which by axiom (iii) implies $x=e\odot x$.

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Yes, $e$ ($=e\odot e$) is a right identity in $\mathcal S$ because $$(x\odot e) \odot e = x\odot (e\odot e)= x\odot e$$ (and is a two sided identity because $\mathcal H$ was abelian to begin with).

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  • $\begingroup$ I changed my preferred answer to Eric Wofsey's because he showed that $e$ is an identity element of $\mathcal{H}$, which obviates the rest of my question. $\endgroup$ – Steven Thomas Hatton Apr 24 at 23:08

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