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Let $L=\{*\}$. The usual axiomatization of groups in this language has the EA axiom $\exists{e}\forall{x}$ $ e*x = x$. But the union of a chain of groups is also a group. This means that the theory of groups has an AE axiomatization. My question is, what is a system of AE axioms for groups.

I want to replace the above axiom with $\exists{y}$ ${y^2=y}$. But I have had trouble showing that the axiom system leads to the theory of groups.

2 Answers2

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Say you add the axiom $\exists e\,\, e^2=e$ and also cancellation laws $ab=ac\implies b=c$ and $ba=ca\implies b=c$. Then $xe^2=xe$, which implies $xe=x$ by cancellation.

David C. Ullrich
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  • +1. Just to be perfectly explicit: note that the cancellation laws are of the form "$\forall \overline{x}[$quantifier-free stuff$]$." – Noah Schweber Jan 14 '16 at 15:39
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Another cute axiomatization is

  • $\exists x\, x = x$ (just to rule out the empty structure, if you allow it)
  • $\forall x\,\forall y\,\forall z\,(xy)z = x(yz)$
  • $\forall x\,\forall y\,\exists z\, xz = y$
  • $\forall x\,\forall y\,\exists z\, zx = y$

It's a bit more complicated to show that every model of these axioms is a group, see this question.

Community
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Alex Kruckman
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  • Thanks Alex. Out of curiosity, did you have an answer to my question about dividing? –  Jan 14 '16 at 21:08