In The Number System by Thurston the author introduces an algebraic structure he calls a "hemigroup". It doesn't appear to be a very common usage. The laws of a hemigroup are:

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

Apparently there are other definitions for the same term. Is there a more common term than "hemigroup" for this kind of structure?

  • 3
    $\begingroup$ Typo in (ii)? [And a small typo in (iv)]. $\endgroup$ – M. Vinay Apr 11 at 1:17
  • $\begingroup$ Unless my eyes are plying tricks on me, The laws are stated consistently with the book. (i)associative; (ii)commutative; (iii) cancellation; (iv) neutral element. $\endgroup$ – Steven Thomas Hatton Apr 11 at 1:23
  • $\begingroup$ Okay, so commutativity is: $(x*y) = (y*x)$ (you've written $(y*z)$ instead of $(y*x)$, unless it's my eyes that are playing tricks on me!) : D $\endgroup$ – M. Vinay Apr 11 at 1:24
  • 1
    $\begingroup$ I don't know if this helps much, but it appears to be a commutative monoid with identity. $\endgroup$ – Gary Moon Apr 11 at 2:02
  • 2
    $\begingroup$ @GaryMoon monoid means "semigroup with identity", so "monoid with identity" means "semigroup with identity with identity". $\endgroup$ – YCor Apr 11 at 9:21

This is usually called a cancellative commutative monoid. Note that in the presence of (iii), (iv) is equivalent to saying that $e$ is a (left) identity, so these axioms just say you have a commutative associative operation with an identity element (i.e., a commutative monoid) which satisfies the cancellation axiom (iii).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.