I'm trying to describe/define the commutative binary operation on a three-element set which when the operands are the same, gives the same element and when they are different gives the element which is not an operand. So for $\{a,b,c\}$, if we denote the operation $\#$: $$a\#a=a\quad \text{and} \quad a\#b=c \quad\text(etc.)$$

I suppose I could just define it case-by-case, but the notion of "the one that's either both of these or neither of these" is so intuitive and natural that that seems like a clunky solution. It also wouldn't be particularly useful to me. I want to describe it with a notation/definition which will lend itself to arguments regarding member-wise application of the operation on tuples i.e. $$(x_1,y_1,z_1...)\mathbf{\#}(x_2,y_2,z_2,...)=(x_1\#x_2,y_1\#y_2,z_1\#z_2,...)$$ In particular I want to discuss the properties of tuple-chains which, through repeated application of the member-wise operation, transform any given tuple into another. That is, I want to find out, given tuples $\mathbf {s}=(x_s,y_s,z_s,...)$ and $\mathbf{d}=(x_d,y_d,z_d,...)$, and considering $\mathbf{\#}$ to be right-associative, what can I say about the tuples in the sequence $\mathbf{C}=\mathbf{t_0},\mathbf{t_1},..\mathbf{t_n}$ if I know that$$\mathbf{t_0\#t_1\#}...\mathbf{\#t_n\#s}=\mathbf{d}$$

I'm not currently looking for help with that question, as I want to explore it myself for a while, and it's pretty open-ended anyway, but I have some ideas I'm setting out to prove, and I'm having trouble developing a formal definition of $\#$ that is useful in this context.

Any Ideas?

  • $\begingroup$ It feels similar to an external direct sum but I'm not sure if that relates to groups only. $\endgroup$
    – Karl
    Aug 24, 2015 at 22:00
  • $\begingroup$ @Karl the member-wise application works the same way, yes, but I'm really concerned with how to formally define the $\#$ operation itself in a way that naturally extends to the tuple version and can thus be used to reason about it. $\endgroup$ Aug 24, 2015 at 23:07

3 Answers 3


I don't really understand the question, but the structure you've discovered is the unique ITSQ on three elements (well-done!).

In detail:

An idempotent totally-symmetric quasigroup (ITSQ) is a set $X$ together with a binary operation $\#$ such that the following hold:

  • $x\#y = y \#x$
  • $(x\#y)\#y=x$
  • $x\#x=x$.

Now let $A = \{a,b,c\}$ and $\#$ be as defined in your question. It's clear that $(A,\#)$ is an ITSQ. I claim that:

Proposition. If $f : A^2 \rightarrow A$ has the property that $(A,f)$ is an ITSQ, then $f=\#$.

Proof. We know that $$f(a,a) = a \qquad f(b,b) = b \qquad f(c,c) = c$$

If we can show that $f(a,b) =c,$ then by symmetry, this completes the proof.

Assume toward a contradiction that $f(a,b) \neq c$. Then we get two cases:

  • Case 0. $f(a,b) = a$.

  • Case 1. $f(a,b) = b$.

Case 1: Suppose $f(a,b)=b$. Then $f(f(a,b),b)=f(b,b)$. So $a=b$, a contradiction.

Case 0: Similar.

  • $\begingroup$ It's funny, that was actually my initial attempt to describe the operation, but it was suggested to me that those criteria were not sufficient to describe it. I very much appreciate your proof that they are. $\endgroup$ Nov 21, 2016 at 21:50

It's somewhat unclear what your criterion is for an acceptable definition, but if you want something that "looks" mathematical, you can identify your three-member set with the cube roots of unity, $\{1,e^{2\pi i/3},e^{-2\pi i/3}\}=\{1,{-1+\sqrt{-3}\over2},{-1-\sqrt{-3}\over2}\}$ and then, using complex conjugation, define

$$x\#y= \begin{cases} x\quad\text{if }x\overline y=1\\\\ \overline{xy}\quad\text{if }x\overline y\not=1 \end{cases}$$

If you like, you can let $\phi$ be an arbitrary bijection between your set $\{a,b,c\}$ and the cube roots of unity, in which case the definition is

$$x\#y= \begin{cases} x\quad\text{if }\phi(x)\overline{\phi(y)}=1\\\\ \phi^{-1}(\overline{\phi(x)\phi(y)}\quad\text{if }\phi(x)\overline{\phi(y)}\not=1 \end{cases}$$

(There is a tacit theorem here that the definition truly is a binary operation, and is independent of the choice of $\phi$.)

Added later: Here is a math-y alternative that avoids breaking into cases: Identify your three-member set with the unit vectors $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ and then, using the dot and cross product in $\mathbb{R}^3$, define

$$u\#v=(u\cdot v)u+((1,1,1)\cdot(u\times v))u\times v$$

The main trick here is that $u\times v=0$ if $u=v$ and $u\cdot v=0$ is $u\not= v$. The $(1,1,1)\cdot(u\times v)$ is to get the sign right when $u\not=v$.

  • $\begingroup$ Thanks for this. You're right in that I didn't define what I was looking for very well, because I didn't really know exactly what I was looking for. I knew that I wanted to prove statements about the tuple-version of this operation (specifically this one) and I was looking for some way to represent the operation which would provide a sort of "way in" to reasoning about it, by analogy to more familiar constructs. Looking "mathy" wasn't the goal, but is a by-product. $\endgroup$ Nov 15, 2016 at 19:54
  • $\begingroup$ In any case, both of your suggestions have the potential to provide the sort of foothold I was looking for. $\endgroup$ Nov 15, 2016 at 19:56

I just came across this question from last year. If you're still interested, here's a simple solution:

Use the set $\mathbb{Z}_3$ of integers modulo $3$ as your three-element set, and define $$x\,\#\,y=2(x+y) \pmod{3}.$$

By the way, if you liked @BarryCipra's idea of using the cube roots of unity as your three-element set, you can define $x\,\#\,y$ to be $x^2 y^2$ or, equivalently, $\dfrac1{xy}.$ This is isomorphic to the solution using $\mathbb{Z}_3$ in the first part of this answer.

  • $\begingroup$ Thanks. I really like the mod 3 interpretation. That is a helpful way of thinking about it. Much appreciated. $\endgroup$ Nov 28, 2016 at 16:43

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