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There's a harder question lurking behind this question that was just asked. The context is quasigroup theory. A commutative quasigroup can be defined as a set $Q$ together with commutative binary operation $*$ such that for all $a,b \in Q$, there is a unique "solution" $s \in Q$ solving $s*a=b$. We write $b/a$ for the unique such $s$. The linked question (essentially) asks if there exists a commutative quasigroup satisfying the identity $a/b=b/a$. (Yes, for example $\mathbb{Z}/2\mathbb{Z}$ has this property with respect to addition.) What I'd like to know is, can we usefully characterize all commutative quasigroups satisfying this identity, including the non-associative ones?

Ideas, anyone?

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goblin GONE
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    Here's a small nonassociative example: $\mathbb Z/3\mathbb Z$ with the operation $ab=2(a+b)$. More generally, $ab=-a-b$ seems to work, starting with an abelian group, and is nonassociative unless all nonzero elements have order 2. – Jonas Meyer Feb 12 '15 at 17:52
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    Apart from elementary $2$ groups (groups where all elements have order dividing$~2$), any (non-empty) examples must be non-associative. Associativity implies having a neutral element, since $a/a$ must be one for every $a$. And then if $e$ is a neutral element then $a=ae=a/e=e/a$ shows that $aa=e$ for all $a$. – Marc van Leeuwen Feb 13 '15 at 13:34
  • Sorry for the banal comment, but maybe this question is hard enough for mathoverflow (no idea when the move is appropriate)? – Blaisorblade Feb 13 '15 at 20:17
  • @JonasMeyer, thanks for the bounty. – goblin GONE Feb 15 '15 at 08:41
  • Note an answer in part to this question that was (unfortunately) given to the other one, involving inverse semigroups (and semilattices): http://math.stackexchange.com/a/1145602/173347 It's not a complete characterization though for what you ask. – the gods from engineering Feb 15 '15 at 14:02
  • One fairly general way to construct these would be to take an abelian group $(G,+)$ and set, for some permutation $f$ on $G$: $$a*b = f(a) - b$$. Then, $b/a = f^{-1}(a+b)$, which is clearly commutative. There are probably other quasigroups not generated this way, though I haven't found any. (@JonasMeyer's example can be built from $(\mathbb Z/3\mathbb Z,+)$ this way) – Milo Brandt Feb 15 '15 at 15:17
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    @Meelo: The operation you defined is usually not commutative. It is if $f(a) = -a$, in which case you get the examples I mentioned. – Jonas Meyer Feb 17 '15 at 15:21
  • @Jonas Meyer: In relation to your example: for any abelian group the subtraction operation (as division) yields a so-called pique, i.e. quasigroup with a "pointed identity" that satisfies only the weaker law $e=ee$. ("Pique" itself is a sort of acronym for "pointed idempotent quasigroup.") This notion of pique hasn't made into Wikipedia, but D.H. Simth's intro book to quasigrups has plenty of results about piques too. – the gods from engineering Feb 23 '15 at 03:30
  • On the matter of terminology, it turns out there's an established name for such a quasigroup: it's not "commutative quasigroup" but rather "totally symmetric quasigroup". Idempotent and totally symmetric quasigroups are (in bijection with) Steiner triples. An interesting [sub]question would be to give an example of a totally symmetric quasigroup that does not have any idempotents. – the gods from engineering Feb 23 '15 at 05:24
  • @JonasMeyer: the fairly general example $a*b=-a-b$ that you found is only a subclass of totally symmetric quasigroups, which are also called "extended triple systems" (ETS); this subclass you found is indeed called entropic or abelian ETS. – the gods from engineering Feb 23 '15 at 06:19
  • The entropic property can be stated as a variety (ab)(cd)=(ac)(bd). Also, what you found is actually a complete characterization of entropic totally symmetric quasigroups, in the sense that every entropic totally symmetric quasigroup is isotopic to exactly one abelian group. – the gods from engineering Feb 23 '15 at 06:44

1 Answers1

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Updating my previously (wrong!) response.

Let $(Q,\cdot)$ be a commutative quasigroup. TFAE:

  1. For all $x,y \in Q$ $xy=x/y$.
  2. For all $x,y \in Q$ $x/y=y/x$.

First note, $(Q,\cdot,\backslash, /)$ is a quasigroup if the following are satisfied for all $x,y \in Q$ $$ x(x\backslash y) = y = x\backslash (xy),\\ (y/x)x = y = (yx)/x. $$ This is an equivalent definition (to unique solutions of $ax=b$ and $ya=b$), but now we are insured that quasigroups form a variety.

Now, it is straightforward to show that if $Q$ is commutative, then $x/y=y\backslash x$. Using this, it is now easy to see that $x/(x/y)=y$ which implies $x*y=x/y$. The other implication is immediate. Again, both directions rely on commutativity.

If you include an identity in your assumptions, ($i.e.$ Q is a loop), then you have that $x=x^{-1}$. So your loop is power associative (it is not necessarily diassociative!). If you add associativity (as already stated), Q is an elementary abelian $2$-group.

Mark Greer
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    As some one who is new to math.stackexchange, I hope the above is on the correct path to your desired answer. One question: What or where is this questions coming from? As a "loop/quasigroup theorist", I'm always interested when people are thinking about these things! – Mark Greer Feb 16 '15 at 17:54
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    I think its cool that you're a loop/quasigroup theorist; out of curiosity, what problems or ideas got you involved in the field? My motivation for this question was pure curiosity; this question made no mention of associativity, and it got me wondering about the non-associative case. – goblin GONE Feb 16 '15 at 19:31
  • Ah, I made a grave mistake! Never try to answer a problem before noon! I'll adjust my answer above (I assume that math.stackexchange will allow this?). – Mark Greer Feb 17 '15 at 00:30
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    @goblin As someone who went to graduate school recently (graduated $2$ years ago), I really enjoyed Group Theory but quickly realized most of the work has already been done. Nonassociative stuff pops up often with lots of constructions, theorems, etc. if you remove a hypothesis or two. Loop/Quasigroups is still a very young subject, so there are lots of interesting problems to work on. However, as my first response was clearly wrong, you can quickly make an easy error and say terrible things! – Mark Greer Feb 17 '15 at 00:53
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    @Meelo I can't seem to comment in the above section, so I apologize for adding this here. In your answer, do you mean to say there exists a permutation $f$? In your example, if I let $f$ be the identity permutation, then $ab=a-b\neq b-a=ba$? – Mark Greer Feb 17 '15 at 01:07
  • @MarkGreer Ah, you're right, I think; I parsed the question as defining a commutative quasigroup to be one in which $b/a=a/b$ (which the quasigroup I define satisfies, regardless of the fact that $ab\neq ba$), but reading more carefully, it looks like this was not the intended meaning. – Milo Brandt Feb 17 '15 at 23:50
  • What if we only assume that $Q$ is a pique instead of a loop? Do we get anything interesting? – the gods from engineering Feb 23 '15 at 03:31
  • @Meelo: The terminology introduced/defined in this question is a bit unfortunate. D.H. Smith's book doesn't define a notion of "commutative quasigroup", but he does define "abelian quasigroup" to be a quasigroup with commutative and associative multiplication. – the gods from engineering Feb 23 '15 at 03:56
  • Also, D.H. Smith defines a "semisymmetric quasigrup" as a quasigrup in which $y/x = xy$ (note the order reversal). I actually wonder if that's not the same thing as what's being talked about here given that there are two ways to define equational division from combinatorial division. – the gods from engineering Feb 23 '15 at 04:01
  • What's perhaps more interesting is that given any quasigroup you can construct a semisymmetric quasigrup from it. – the gods from engineering Feb 23 '15 at 04:06
  • Ah, never mind, the opposite division requires opposite multiplication, so this notion is not the same as semisymmetric quasigrup. D.H. Smith calls what is being talked about in this question/answer "totally symmetric quasigroup", i.e. $xy = x\backslash y = x/y$. – the gods from engineering Feb 23 '15 at 04:10
  • There's a bijection between idempotent, totally symmetric quasigroups and Steiner triples. – the gods from engineering Feb 23 '15 at 04:29
  • Some more interesting bits from Smith's book: For any element $e$ of a totally symmetric quasigroup $(Q; \cdot)$, the binary operation defined by $x + y = exy$ forms a loop with the identity element $e$; notation: $F_e(Q; \cdot) = (Q;+; e)$. Furthermore, there's not much agreement in the literature what an "abelian quasigroup" should denote. Smith cites Manin as defining "abelian quasigroup" as a totally symmetric quasigroup with the additional property that for each element $e \in Q$, the loop $F_e(Q;\cdot)$ must be an abelian group. – the gods from engineering Feb 23 '15 at 04:46
  • Manin's motivation for his definition of abelian quasigroup (found in his book Cubic Forms) is an interesting example of a totally symmetric quasigroup; in Smith's presentation: "Let $V$ be an irreducible cubic curve in the complex projective plane $\mathrm{PG}(2;\mathbb{C})$. Let $Q$ be the set of simple points of $V$. Specify the ternary multiplication table of a quasigroup structure $(Q; \cdot)$ on $Q$ to consist of collinear triples $(x; y; z)$. If two of $x, y, z$ coincide, then the line on which they lie is tangent to $V$. All three coincide if and only if $x$ is a flex of $V$." – the gods from engineering Feb 23 '15 at 05:03
  • Finally, if you do not restrict the cubic curve to be embedded in a projective plane (like Manin did), then the totally symmetric quasigroup (no longer just Abelian in Manin's sense) corresponds to Buekenhout's generalized elliptic cubic curve (GECC). – the gods from engineering Feb 23 '15 at 05:40
  • Nice answer, deserves more upvotes. Also, I think it would be good to fill in some of the algebraic details, because those unused to quasigroup theory may have difficulty proving the statements themselves. – goblin GONE Mar 17 '15 at 14:47
  • @goblin: they are in Smith's book. They go pretty far back too, at least to the 1975 paper by Lindner and Steedley "On the number of conjugates of a quasigroup". – the gods from engineering Apr 07 '15 at 13:09
  • [It looks like MS.SE crashed as I was editing my previous post. I had to wait for it become usable again. Anyway...] Of particular interest in the aforementioned paper is Corollary 5. The case of the totally symmetric quasigroup was known earlier though. See Theorem 8.3 in Stein's 1957 "Foundations" paper. – the gods from engineering Apr 07 '15 at 13:42