Consider a quasigroup $(S,+)$ such that for every $a,b,c,d\in S$, $$(a+(b+c))+d=a+(b+(c+d)).$$

This is almost a group, but not quite. For instance, $(\mathbb Z,-)$ satisfies those axioms.

You can easily prove that for any element $x$, the operation $a+_xb:=a+(x+b)$ yields a group.

But I wonder: for any given $(S,+)$ and any pair $x,y \in S$, are $(S,+_x)$ and $(S,+_y)$ neccessarily isomorphic?

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    $\begingroup$ In fact, $(G,-)$ satisfies these axioms for $G$ any abelian group, where $-$ is interpreted as $(a,b)\mapsto (a+(-b)$. $\endgroup$ Aug 6, 2020 at 12:11
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    $\begingroup$ I don't know if this is useful, but I've just noticed that $a+_x (b +_y c) = (a +_x b) +_y c$ holds $\endgroup$ Aug 6, 2020 at 13:20
  • $\begingroup$ If $(S,-)$ is obtained from an abelian group $(G,+)$ by the above procedure, then I claim that the group $(S,+_x)$ is isomorphic to $G=(S,+_0)$. I believe the function $a\mapsto a+x$ is a homomorphism of groups. $\endgroup$ Aug 6, 2020 at 13:25
  • $\begingroup$ Well, in this specific case it's obvious. But I believe you can have quasigroups that still satisfy this and aren't in this form. $\endgroup$ Aug 6, 2020 at 13:33
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    $\begingroup$ To be fair, I'm not much of anything theorist, just a student. I don't know if this specific result appears in any literature, it just seemed fairly obvious to me. But you're right, I should have said so in the question, my bad. I'm terribly sorry four your waiste of time (if you do indeed treat it as such). $\endgroup$ Aug 6, 2020 at 13:54

1 Answer 1


A colleague of mine has been able to prove that this is, indeed, the case. His proof had some redundant steps, below is a polished version.

Theorem: If a quasigroup $(S,+)$ satisfies the axiom $(*)$:

$$ (a+(b+c))+d=a+(b+(c+d))$$

Then for any $x \in S$, operation $a+_xb:=a+(x+b)$ forms a group. Furthermore, any two such groups $(S,+_x)$,$(S,+_y)$ for given $(S,+)$ are isomorphic.

Lemma 1: $(S,+)$ has right-identity.

Proof: $(S,+)$ is a quasigroup, for any $c$ there exists $d$ such that $c+d=c$. Therefore:


Let $x:=(a+(b+c))$. Since we've put no constraints on $a,b,c$, this can be any element of $S$. We have:


So $d$ is the right-identity. From this point on the right-identity will be denoted as $0$.

Corollary: From the existence of right-identity and the properties of a quasigroup, it trivially follows that any operation $+_x$ has a right-identity and right-invertability.

Lemma 2: $+_x$ is asociative.


$(a+_xb)+_xc\overset{def}{=}(a+(x+b))+(x+c)\overset{(*)}{=} a+(x+(b+(x+c)))\overset{def}{=}a+_x(b+_xc)$

This establishes that $(S,+_x)$ indeed forms a group.

Lemma 3: $$0+(0+a)=a$$

Proof: $$x+a=(x+(0+0))+a\overset{(*)}{=}x+(0+(0+a))$$

Since $(S,+)$ is a quasigroup, the lemma follows.

Lemma 4: $$(x+y)+(0+z)=x+(y+z)$$



Now, we have sufficient tools to prove that $f(x)=k+x$ is a homomorphism from $(S,+_k)$ to $(S,+_0)$ for any $k \in S$


$$\cdots \overset{(*)}{=} k+(x+(k+y))\overset{def}{=}f(x+_ky)$$

This establishes homomorphism. Now, since $(S,+)$ is a quasigroup, $f(x)$ a bijection and therefore isomorphism. Since for any two $x,y \in S$, $(S,+_x)$ and $(S,+_y)$ are isomorphic with $(S,+_0)$ and isomorphism is obviously transitive, this proves the theorem.

Corollary: this theorem led to a surprizing rezult. All quasigroups $(S,+)$ satisfying the axiom $(*)$ can be represented as: $x+y=x \star f(y)$, where $(S,\star)$ is a group, and where $f(x)$ is an automorphism of said group such that $f(f(x))=x$. Furthermore, any operation $x \star f(y)$ forms a quasigroup satisfying the axiom $(*)$. The proof isn't hard, but it's beyond the scope of the original question, so I'll omit it.


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