# Do Any Other 3-element Structures With a Single Binary Operation Exist Which Satisfy This Axiom Set?

Let $x$ and $y$ denote variables over the set $\{a, b, c\}=S$, and $t$ some constant of $S$. Let "X" denote an unknown binary operation on $S$. Suppose we have the following set of properties:

1. For all x, xxX=t
2. There exists a y, such that for all x, xyX=y=t (the same constant as in 1.)
3. There exists an x, such that for all y, xyX=t
4. There exists an x, such that for all y, xyX=y
5. There exists an x, such that there exists a y not equal to x, and xyX=x.

There exist at least 6 structures (all automorphic to each other, which indicates how I obtained them in the first place, THEN I wrote the properties) which satisfy this set of properties as follows:

A  a  b  c
a  c  c  c
b  b  c  c
c  a  b  c

B  a  b  c
a  b  b  b
b  a  b  c
c  c  b  b

C  a  b  c
a  c  a  c
b  c  c  c
c  a  b  c

D  a  b  c
a  a  b  c
b  a  a  a
c  a  c  a

E  a  b  c
a  b  b  a
b  a  b  c
c  b  b  b

F  a  b  c
a  a  b  c
b  a  a  b
c  a  a  a


Do any other binary operations on {a, b, c} satisfy properties 1.-5. above? I would think "yes", since, if I've gotten things right, property 1. specifies something about the diagonal, property 2. specifies a particular column, for which all of its values equal that of the diagonal, 3. specifies a row that always equals the valued of the diagonal, 4. specifies a row which always equals the values of the second coordinate, and I think 5. specifies the only value left undetermined by 1.-4. Do any other structures exist here?

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@Doug: Your complaint on "needs more symbols" simply means that you value your time (the insignificant amount that it would take to type extra prentheses), while at the same time finding other people's time not worth your concern (the time it takes them to figure out what you mean); in short, contempt for others while you ask for their help. As they say in Mexico, "limosnero con garrote." Given that my time is so worthless to you, this was my last effort on your behalf. –  Arturo Magidin Jun 21 '11 at 4:28
@Doug: In short, if people want to talk to you, they need to do it in whatever way you prefer. Like I said, limosnero con garrote. Enjoy your monologues, Doug; please don't address any more comments to me. –  Arturo Magidin Jun 21 '11 at 4:44
-1 for purposefully bad notation. How could you expect anyone to be able to read this question? –  Jim Belk Jun 21 '11 at 5:27
@Doug: as a passerby to this conversation, let me say that your most recent comment strikes me as simply obnoxious. You have decided that you wish to communicate with a hypothetical audience rather than the audience that you actually have (most of the world -- or more! -- learns mathematics using "infix notation"). Arturo, Jim and Mariano are all experienced mathematicians and talented expositors and communicators of mathematics: they know what they are talking about. Let me also say that I was one of the ones who stopped thinking about your question because of the unfriendly notation. –  Pete L. Clark Jun 21 '11 at 18:33
@Doug: Sigh. Puesto que hay mas hispanohablantes como lengua nativa que angloparlantes, obviamente debemos comunicarnos en español, pues eso requiere de menos suposiciones. O quizas en Mandarin. En cualquier caso, Sr Maderacuchara, que tenga usted buen dia, buena semana, buena decada, y buena vida. Estoy harto de sus argumentos de espantapajaro y su malrepresentacion de sus propias palabras en el servicio de pretender que usted no dijo las absurdeces que ya dijo, una y otra vez, por no decir cuando trata de meter palabras en mi boca o pensamientos en mi cerebro. Que se divierta, pero solito. –  Arturo Magidin Jun 22 '11 at 3:02

So, suppose you have a 3 element magma $M$, and an element $t$ such that:

1. For all $x\in M$, $xx=t$.
2. There exists $y\in M$ such that for all $x\in M$, $xy=t$.
3. There exists $w\in M$ such that for all $x\in M$, $wx=t$.
4. There exists $z\in M$ such that for all $x\in M$, $zx=x$.
5. There exist $x,y\in M$, $x\neq y$, such that $xy=x$.

Note that $z\neq w$: letting $x\neq t$ (possible since $M$ has three elements), then $wx = t \neq x = zx$ by 3 and 4, so $z\neq w$.

Note that $z=t$: for $t=zz=z$, by 1 and 4.

Likewise, $y=t$, since $t = zy = y$ (by 2 and 4). Thus, $y=z=t$, and $w$ is different from $t$.

Thus, $t$ is a right zero and a left identity. So we have:

$M$ is a 3-element magma; there are distinguished elements $t$ and $w$ such that:

1. $t\neq w$.
2. $xx = t$ for all $x\in M$.
3. $xt = t$ for all $x\in M$.
4. $tx = x$ for all $x\in M$.
5. $wx = t$ for all $x\in M$.
6. There exists $z,r\in M$, $z\neq r$, such that $zr=z$.

From 4 and the clause $z\neq r$ in 6, we have that $z\neq t$. From this and 5, we have that $z\neq w$. Thus, the three elements of $M$ are $t$, $w$, and $z$. From 3, we know that $r\neq t$, and since $r\neq z$, hence $r=w$. Thus:

$M$ is a three element magma. The three elements are $t$, $w$, and $z$. They satisfy:

• $xx = t$ for all $x$.
• $xt = t$ for all $x$.
• $tx = x$ for all $x$.
• $wx = t$ for all $x$.
• $zw = z$.

Thus, the multiplication table of $M$ is: $$\begin{array}{c|ccc} &t&w&z\\ \hline t&t&w&z\\ w&t&t&t\\ z&t&z&t \end{array}$$ The first row is forced by the third listed property in the last quote box. The first column is forced the second listed property; the second row by the fourth listed property; the last diagonal entry by the first listed property; and the $zw$ entry by the last listed property. This gives all the entries of the table.

Thus, the properties completely determines $M$ up to the names of the elements. If you want to call the elements $a$, $b$, and $c$, then there are exactly $3!=6$ ways in which you can assign the names $a$, $b$, and $c$ to $t$, $w$, and $z$, so that gives exactly six different tables in terms of $a$, $b$, and $c$, though they are all isomorphic to $M$.

(And it would have taken me about half as long to figure out what you are asking if you were using standard notations and nomneclature instead of insisting on your own personal language; I wonder if your favorite movie is Nell.)

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Thanks! –  Doug Spoonwood Jun 21 '11 at 5:18