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I did a project on compilation of few results in reverse mathematics few weeks back and had given an example of Group Axioms stating that those are independent of each other i.e. we can get a well-defined structure even though any one or two of the axioms aren't considered. For instance, the loop $(\mathbb Z, -)$ which has identity & invertibility but no associativity. Or a monoid which has associativity & identity but no invertibility.

However, yesterday on the wiki page of quasigroups I read "A nonempty associative quasigroup is a group" with this proof.

Earlier, I'd checked these two answers on SE [1] [2] and arrived at my previous conclusion. Also I read this paper by Jacobson & Yocom strengthening the belief.

But I'm a bit confused now after reading the proof for associative quasigroup. Am I missing some detail which differentiates invertibility from divisibility?

Parth Bhagwat
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1 Answers1

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An associative quasigroup doesn't have to be a group, but a nonempty one does. So a group is an associative quasigroup with one more axiom. You may take this axiom to be "there's an identity", or (as the linked proof shows) you can weaken it to "there's an element".

Divisibility is defined without an identity as the Latin square property: for each $a,\,b$ there exist unique $x,\,y$ with $ax=ya=b$. This also implies a quasigroup is cancellative, i.e. $ab=ac\implies b=c$. The linked proof uses this to show a nonempty associative quasigroup will have an identity.

J.G.
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  • So, the group axioms aren't independent? Are the answers to the 2 linked MSE questions incorrect then? – Parth Bhagwat Feb 16 '25 at 13:59
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    @ParthBhagwat The group axioms are independent. However, there's more than one choice of axioms. (Theories tend to have multiple equivalent axiomatisations.) You can say it's an associative quasigroup with an identity, or you can say it's an associative quasigroup with at least one element. – J.G. Feb 16 '25 at 14:06
  • In fact you do not need specify the cancellative property: Any non-empty set with associative product satisfying for all $a,b$ there exist $x,y$ with $ax=ya=b$, will be a group. That is, there is no need to demand that $x,y$ are unique. Pf: There is some element $x$. There is some element $u$ satisfying $ux=x$. For any $y$, there is some $z$ satisfying $y=xz$. Thus $uy=uxz=xz=y$ and $u$ is a left identity. Similarly we have a right identity $v$, and $u=uv=v$ is a 2-sided identity. Then for any element $x$, we have $y,z$ with $xy=u, zx=u$ and $y=zxy=z$ is the inverse of $x$. – tkf Feb 17 '25 at 04:42
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    @tkf Good to know. It's just not the strategy Planet Math used. – J.G. Feb 17 '25 at 08:08