# Does this almost-group uniquely define a group?

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?

• In fact, $(G,-)$ satisfies these axioms for $G$ any abelian group, where $-$ is interpreted as $(a,b)\mapsto (a+(-b)$. Aug 6, 2020 at 12:11
• 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 Aug 6, 2020 at 13:20
• 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. Aug 6, 2020 at 13:25
• 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. Aug 6, 2020 at 13:33
• 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). Aug 6, 2020 at 13:54

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:

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

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

$$x+d=x$$

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.

Proof:

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

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

Proof:

$$(x+y)+(0+z)=(x+(y+0))+(0+z)\overset{(*)}{=}x+(y+(0+(0+z)))\overset{\text{L3}}{=}x+(y+z)$$

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

$$f(x)+_0f(y)\overset{def}{=}(k+x)+(0+(k+y))\overset{(*)}{=}((k+x)+(0+k))+y\overset{\text{L4}}{=}(k+(x+k))+y\overset{(*)}{=}\cdots$$

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

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.