Recall that a quasigroup is a pair $(Q, \ast)$, where $Q$ is a set and $\ast$ is a binary product $$\ast: Q \times Q \to Q$$ satisfying the Latin square property, namely that for all $x, y \in Q$ there is a unique $a \in Q$ such that $y = ax$ and a unique $b \in Q$ such that $y = x b$, or equivalently, that the multiplication table of $\ast$ is a Latin square. A quasigroup $(Q, \ast)$ is a loop iff it has an identity element, that is an element $1$ such that $1\ast x = x = x\ast 1$ for all $x \in Q$.

This question asks about constructing a loop $(L, \ast)$ on a set $L$ of five elements (denote $L = \{1, a, b, c, d\}$) that satisfies the involution condition $x^2 = 1$ for all $x \in L$. My answer there shows that there is only one such loop up to isomorphism; its multiplication table is: $$ \begin{array}{c|ccccc} \ast & 1 & a & b & c & d \\ \hline 1 & 1 & a & b & c & d \\ a & a & 1 & c & d & b \\ b & b & d & 1 & a & c \\ c & c & b & d & 1 & a \\ d & d & c & a & b & 1 \end{array}.$$ (Note that it is nonassociative, as $(ab)d = a \neq ac = a(bd)$.)

Even without the involution condition, this example is minimal in the sense that any loop of order $< 5$ is in fact a group. (Up to isomorphism there are six loops of order $5$: This one, the group $(\mathbb{Z}_5, +)$, and four other non-groups.)

So, given that this example is both minimal and unique, it's natural to ask:

Is there a more informative/interesting way to view the loop structure $\ast$ on $L$ than via its multiplication table? That is, does it arise naturally in some other setting?

  • 2
    $\begingroup$ It's an excellent question! I think the real problem comes in the combination of non-associativity along with the (partial) asymmetry; non-associativity can be handled by e.g. operations like $*(a,b)= a-b$ (or roughly-equivalently, $*(a,b)=\frac ab$), but then the existence of a two-sided identity forces something more like $*(a,b)=|a-b|$ - but that doesn't work with the asymmetry for non-identity operations. $\endgroup$ – Steven Stadnicki Apr 2 '15 at 15:29

There are 4 quasi-quaternion groups, strictly $abc, adb, acd, bdc$, (even, odd, even, odd permutations of 3 of $abcd$) with $xy=z, yz=x, zx=y, (xy)z=x(yz)=1$.

They can be visualised on 4 tetrahedral directed graphs, with $x\to y$ implying we follow the arrow, otherwise follow the blank edge.

Quasi-group order 5 on 4 tetrahedral directed graphs

These can be combined into a tetrahedron with each face given a clockwise or anti-clockwise spin (all faces have the same spin in this case - viewed from outside the tetrahedron). Adding the identity involves adding bidirectional edges and loops.

The idea can be extended to other solids with 3 edges per vertex, e.g. the cube and the dodecahedron.

| cite | improve this answer | |
  • $\begingroup$ This is quite a nice interpretation, cheers, Jon! $\endgroup$ – Travis Willse Apr 12 '15 at 12:43

A Loop with the involution condition you describe has $1$'s along it's main diagonal. Lets remove the $1$ from the Loop to define a related Quasigroup. We maintain the rest of the structure, but need to fill in the now empty entries on the main diagonal. Each row and column are missing a single element and there is only way to satisfy the Latin square condition: the new Quasigroup must be idempotent! Note that we can do the reverse process uniquely too, so there is a one to one correspondence between Loops with this involution condition and Idempotent Quasigroups, so this construction is not so arbitrary.

$\begin{array}{c|ccccc} \ast & 1 & a & b & c & d \\ \hline 1 & 1 & a & b & c & d \\ a & a & 1 & c & d & b \\ b & b & d & 1 & a & c \\ c & c & b & d & 1 & a \\ d & d & c & a & b & 1 \end{array} \Leftrightarrow \begin{array}{c|cccc} ? & a & b & c & d \\ \hline a & & c & d & b \\ b & d & & a & c \\ c & b & d & & a \\ d & c & a & b & \end{array} \Leftrightarrow \begin{array}{c|cccc} \lhd & a & b & c & d \\ \hline a & a & c & d & b \\ b & d & b & a & c \\ c & b & d & c & a \\ d & c & a & b & d \end{array}$

Now that we've done that, the Quasigroup we've constructed has a nice interpretation.

Consider a conjugacy class of elements of order $3$ from the Alternating Group on a set of $4$ elements, $A_4$. Say, $\{(1,2,3),(1,4,2),(1,3,4),(2,4,3)\}$. Define the operation $x\lhd y = xyx^{-1}$ where the implied operations on the right hand side are the usual multiplication of permutations, that is, our operation now conjugates elements.






and we get our quasigroup!

For example $a\lhd b=(1,2,3)\lhd(1,4,2) = (1,2,3)(1,4,2)(1,2,3)^{-1}=(1,2,3)(1,4,2)(1,3,2)=(2,4,3)=c$

See also, conjugation and Quandles.

| cite | improve this answer | |
  • $\begingroup$ This is a nice description! $\endgroup$ – Travis Willse Dec 28 '18 at 18:36

We can start by defining $ab=c$ and also note that if we have the four distinct elements as $\{w,x,y,z\}$, then if $wx=y, xw=z$, that is $xw$ is the only element unused. We also have the property that:

$$ab=c, bc=a, ca=b.$$

Using the 'opposite' rule, we therefore have:

$$ba=d, cb=d, ac=d.$$

We can now deduce the remaining operations, using the definition of a Latin Square.

$ad=b$ (from $ab=c$ and $ac=d$), so also $bd=c$ and $cd=a.$

with the corresponding opposites:

$$da=c, db=a, dc=b.$$

There are a few other points of interest: $(xy)x=y$, $(ab)(ca)=(cb)$, and if $y\ne z$ then $(xy)z=y(zx)$.

I wrote a JavaScript program, which originally was going to parse expressions from $L$, but at the moment just displays the multiplication tables for 3 elements.

<!DOCTYPE html>
<title>Quasigroup Explorer</title>
span {
#input {
#output {
#parse {
font:24pt tahoma;
<body onkeydown='span();'>
<span id='input' ></span>
<span id='output' ></span>
<span id='parse' onclick='parse();'>parse</span>

mult=new Array;

function span(e) {
var key=event.keyCode;
if (key!=48 && key!=49 && key!=57 && key!=65 && key!=66 && key!=67 && key!=68 && key!=32) return;
var inp=input.textContent;
var chr='';
if (key=='48') chr=')';
if (key=='49') chr='1';
if (key=='57') chr='(';
if (key=='65') chr='a';
if (key=='66') chr='b';
if (key=='67') chr='c';
if (key=='68') chr='d';
if (key==32 && inp.length>0)  inp=inp.slice(0,-1);

function parse() {
var brk=0;
var inp=input.textContent;
var inpl=inp.length;
var i;
for (i=0;i<inpl;i++) if (inp[i]=='(') brk++;

for (a=0;a<5;a++)
for (b=0;b<5;b++)
for (c=0;c<5;c++) {

/* original
for (a=0;a<5;a++)
for (b=0;b<5;b++)
for (c=0;c<5;c++) {
if (x==0) c0++;
if (x==4) c4++;

| cite | improve this answer | |
  • $\begingroup$ Thanks for the answer, but this doesn't appear to address the question---perhaps I'm missing something?. I understand how to produce the multiplication table from the Latin square rule and a choice like $ab = c$---in fact, this is how I produced the multiplication table in the first place (there are some more details in my answer to the linked question). $\endgroup$ – Travis Willse Apr 8 '15 at 12:57
  • $\begingroup$ @Travis;' Is there a more informative/interesting way to view the loop structure ∗ on L than via its multiplication table? That is, does it arise naturally in some other setting?' - as far as I can see the answer is 'No', but I'm working on it. $\endgroup$ – JMP Apr 8 '15 at 12:59
  • $\begingroup$ My question is whether there's a more useful way to think about this structure than just as some product defined by a particular multiplication table. For example, one can write down the multiplication table for the group $S_3$, but this is not a particularly informative description of the group. It's much more illuminating to think of $S_3$ as the group of permutations of three objects, or as the group of symmetries of an equilateral triangle, or the collection of $3 \times 3$ matrices with one $1$ in each row and column and all other entries zero (under multiplication). $\endgroup$ – Travis Willse Apr 8 '15 at 13:00
  • $\begingroup$ Cheers, Jon, I look forward to seeing whatever you come up with, even if it's negative! $\endgroup$ – Travis Willse Apr 8 '15 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.