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I have been reading The Number System by Thurston. In this book he introduces a thing he calls a 'hemigroup' which I think would more conventionally be an 'abelian cancellative monoid'. Hemigroups are $(G,\cdot)$ such that

  • $(x\cdot y)\cdot z = x \cdot (y\cdot z)$;
  • $x\cdot y = y\cdot x$;
  • if $x\cdot y = x\cdot z$ then $y = z$;
  • there is $e$ such that $e\cdot e = e$.

Well, it's then easy to show that $e$ is an identity: $x\cdot e = x\cdot(e\cdot e) = (x\cdot e)\cdot e$ and then by cancellativity $x\cdot e = x$.

My question then is: can you introduce an identity for groups the same way? So instead of defining a group as $(G,\cdot)$ where

  • $(x \cdot y)\cdot z = x \cdot (y\cdot z)$;
  • there is an $e$ such that $e\cdot x = x$ for all $x$;
  • For all $x$ there is an $x'$ such that $x'\cdot x = e$;

can you replace the second axiom above by 'there is an $e$ such that $e\cdot e = e$' and derive that $e$ is an identity?

(I think you probably can not, but I'm not sure. Certainly for groups it's easy to show that if $a\cdot a = a$ then $a = e$, but you do this by using that $e$ actually is a (left-)identity. However I am very bad at finding these kind of proofs or showing there are none: I can follow them, but I am terrible at finding new ones.)

ignis volens
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2 Answers2

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Try $\{0,1\}$ under multiplication with $e=0$.

Derek Holt
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  • But ${0,1}$ is not a group under multiplication... – William Aug 30 '22 at 09:33
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    @William: but it does satisfy my alternative axioms: $0\times 0 = 0$ but $0$ completely fails to be an identity. So my alternative axioms are not enough to ensure that $e$ is in fact an identity. – ignis volens Aug 30 '22 at 09:47
  • @ignisvolens Oops. I misread your question. My bad. – William Aug 30 '22 at 09:49
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    Thank you for this: I didn't know which answer to accept: I've accepted the other one because it was slightly more general (and I think the underlying thing is that cancellation is not an axiom or groups) but this one also helped me understand my confusion. So, thank you again. – ignis volens Aug 30 '22 at 09:54
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Note that hemigroups have a built-in axiom for cancellation, but your alternative definition of group does not. You can use this observation to break your alternative definition.

Take your favorite group $G$ and form $A = G \cup \{u\}$ with $u$ some new element. On $A$, define the group operation $\cdot$ as on $G$ and, additionally, $x \cdot u = u \cdot x = u$ for all $x \in A$. This $A$ satisfies your alternative definition (taking $u$ as $e$), but is obviously not a group.

Magdiragdag
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    We can't take the unit of $G$ as $e$ because otherwise $u$ has no inverse (if i'm not mistaken) – 8bc3 457f Aug 30 '22 at 09:48
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    Thank you. In particular (as a generalisation of the other answer) if I add $0$ to the non-zero reals under multiplication, say, then $0\times 0 = 0$ but $0$ completely fails to be an identity and indeed the whole thing is not a group at all. – ignis volens Aug 30 '22 at 09:51
  • @8bc3457f You're right, taking $e_G$ as $e$, there is no $x'$ such that $x' \cdot u = e$. I've repaired the answer. – Magdiragdag Aug 30 '22 at 10:34
  • @ignisvolens Indeed. In fact, every field is a counterexample. – Magdiragdag Aug 30 '22 at 10:36