The unique loop (quasigroup with unit) $L$ of order $5$ satisfying $x^2 = 1$ for all $x in L$
$begingroup$
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 $1ast x = x = xast 1$ 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?
abstract-algebra quasigroups
asked Apr 2 '15 at 6:31
TravisTravis
64.4k769151
$endgroup$
add a comment |
$begingroup$
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 $1ast x = x = xast 1$ 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?
abstract-algebra quasigroups
asked Apr 2 '15 at 6:31
TravisTravis
64.4k769151
$endgroup$
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
add a comment |
$begingroup$
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 $1ast x = x = xast 1$ 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?
abstract-algebra quasigroups
asked Apr 2 '15 at 6:31
TravisTravis
64.4k769151
$endgroup$
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 $1ast x = x = xast 1$ 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?
abstract-algebra quasigroups
abstract-algebra quasigroups
asked Apr 2 '15 at 6:31
TravisTravis
64.4k769151
asked Apr 2 '15 at 6:31
TravisTravis
64.4k769151
edited Dec 27 '18 at 19:27
Travis
asked Apr 2 '15 at 6:31
TravisTravis
64.4k769151
asked Apr 2 '15 at 6:31
TravisTravis
64.4k769151
asked Apr 2 '15 at 6:31
TravisTravis
64.4k769151
64.4k769151
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
add a comment |
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
2
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
$begingroup$
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 $xto y$ implying we follow the arrow, otherwise follow the blank edge.
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.
answered Apr 8 '15 at 21:01
JonMark PerryJonMark Perry
11.3k92438
$endgroup$
$begingroup$
This is quite a nice interpretation, cheers, Jon!
$endgroup$
– Travis
Apr 12 '15 at 12:43
add a comment |
$begingroup$
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 $yne z$ 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>
<html>
<title>Quasigroup Explorer</title>
<style>
span {
width:200px;
height:32pt;
white-space:pre;
display:block;
}
#input {
background-color:red;
}
#output {
background-color:blue;
}
#parse {
text-align:center;
font:24pt tahoma;
background-color:green;
}
</style>
<body onkeydown='span();'>
<span id='input' ></span>
<span id='output' ></span>
<span id='parse' onclick='parse();'>parse</span>
</body>
<script>
mult=new Array;
mult[0]=[0,1,2,3,4];
mult[1]=[1,0,3,4,2];
mult[2]=[2,4,0,1,3];
mult[3]=[3,2,4,0,1];
mult[4]=[4,3,1,2,0];
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);
inp+=chr;
input.textContent=inp;
}
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++;
}
out='';
c0=0;c4=0;
for (a=0;a<5;a++)
for (b=0;b<5;b++)
for (c=0;c<5;c++) {
x=mult[mult[a][b]][c];
out+='('+a+''+b+')'+c+'='+x+'<br>';
x=mult[b][mult[c][a]];
out+=b+'('+c+''+a+')='+x+'<br>';
}
output.innerHTML=out;
/* original
out='';
c0=0;c4=0;
for (a=0;a<5;a++)
for (b=0;b<5;b++)
for (c=0;c<5;c++) {
x=mult[mult[a][b]][c];
out+='('+a+''+b+')'+c+'='+x+'<br>';
if (x==0) c0++;
if (x==4) c4++;
x=mult[a][mult[b][c]];
out+=a+'('+b+''+c+')='+x+'<br>';
}
output.innerHTML=out;
*/
</script>
</html>
answered Apr 7 '15 at 13:50
JonMark PerryJonMark Perry
11.3k92438
$endgroup$
$begingroup$
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
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.
$endgroup$
– JonMark Perry
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
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
Apr 8 '15 at 13:01
add a comment |
$begingroup$
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 $xlhd y = xyx^{-1}$ where the implied operations on the right hand side are the usual multiplication of permutations, that is, our operation now conjugates elements.
Let
$arightarrow(1,2,3)$
$brightarrow(1,4,2)$
$crightarrow(2,4,3)$
$drightarrow(1,3,4)$
and we get our quasigroup!
For example $alhd b=(1,2,3)lhd(1,4,2) = (1,2,3)(1,4,2)(1,2,3)^{-1}=(1,2,3)(1,4,2)(1,3,2)=(2,4,3)=c$
See also, conjugation and Quandles.
answered Dec 28 '18 at 17:04
Josh B.Josh B.
2,55511425
$endgroup$
$begingroup$
This is a nice description!
$endgroup$
– Travis
Dec 28 '18 at 18:36
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
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 $xto y$ implying we follow the arrow, otherwise follow the blank edge.
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.
answered Apr 8 '15 at 21:01
JonMark PerryJonMark Perry
11.3k92438
$endgroup$
$begingroup$
This is quite a nice interpretation, cheers, Jon!
$endgroup$
– Travis
Apr 12 '15 at 12:43
add a comment |
$begingroup$
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 $xto y$ implying we follow the arrow, otherwise follow the blank edge.
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.
answered Apr 8 '15 at 21:01
JonMark PerryJonMark Perry
11.3k92438
$endgroup$
$begingroup$
This is quite a nice interpretation, cheers, Jon!
$endgroup$
– Travis
Apr 12 '15 at 12:43
add a comment |
$begingroup$
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 $xto y$ implying we follow the arrow, otherwise follow the blank edge.
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.
answered Apr 8 '15 at 21:01
JonMark PerryJonMark Perry
11.3k92438
$endgroup$
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 $xto y$ implying we follow the arrow, otherwise follow the blank edge.
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.
answered Apr 8 '15 at 21:01
JonMark PerryJonMark Perry
11.3k92438
edited Apr 11 '15 at 4:54
answered Apr 8 '15 at 21:01
JonMark PerryJonMark Perry
11.3k92438
answered Apr 8 '15 at 21:01
JonMark PerryJonMark Perry
11.3k92438
answered Apr 8 '15 at 21:01
JonMark PerryJonMark Perry
11.3k92438
11.3k92438
$begingroup$
This is quite a nice interpretation, cheers, Jon!
$endgroup$
– Travis
Apr 12 '15 at 12:43
add a comment |
$begingroup$
This is quite a nice interpretation, cheers, Jon!
$endgroup$
– Travis
Apr 12 '15 at 12:43
$begingroup$
This is quite a nice interpretation, cheers, Jon!
$endgroup$
– Travis
Apr 12 '15 at 12:43
$begingroup$
This is quite a nice interpretation, cheers, Jon!
$endgroup$
– Travis
Apr 12 '15 at 12:43
add a comment |
$begingroup$
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 $yne z$ 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>
<html>
<title>Quasigroup Explorer</title>
<style>
span {
width:200px;
height:32pt;
white-space:pre;
display:block;
}
#input {
background-color:red;
}
#output {
background-color:blue;
}
#parse {
text-align:center;
font:24pt tahoma;
background-color:green;
}
</style>
<body onkeydown='span();'>
<span id='input' ></span>
<span id='output' ></span>
<span id='parse' onclick='parse();'>parse</span>
</body>
<script>
mult=new Array;
mult[0]=[0,1,2,3,4];
mult[1]=[1,0,3,4,2];
mult[2]=[2,4,0,1,3];
mult[3]=[3,2,4,0,1];
mult[4]=[4,3,1,2,0];
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);
inp+=chr;
input.textContent=inp;
}
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++;
}
out='';
c0=0;c4=0;
for (a=0;a<5;a++)
for (b=0;b<5;b++)
for (c=0;c<5;c++) {
x=mult[mult[a][b]][c];
out+='('+a+''+b+')'+c+'='+x+'<br>';
x=mult[b][mult[c][a]];
out+=b+'('+c+''+a+')='+x+'<br>';
}
output.innerHTML=out;
/* original
out='';
c0=0;c4=0;
for (a=0;a<5;a++)
for (b=0;b<5;b++)
for (c=0;c<5;c++) {
x=mult[mult[a][b]][c];
out+='('+a+''+b+')'+c+'='+x+'<br>';
if (x==0) c0++;
if (x==4) c4++;
x=mult[a][mult[b][c]];
out+=a+'('+b+''+c+')='+x+'<br>';
}
output.innerHTML=out;
*/
</script>
</html>
answered Apr 7 '15 at 13:50
JonMark PerryJonMark Perry
11.3k92438
$endgroup$
$begingroup$
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
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.
$endgroup$
– JonMark Perry
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
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
Apr 8 '15 at 13:01
add a comment |
$begingroup$
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 $yne z$ 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>
<html>
<title>Quasigroup Explorer</title>
<style>
span {
width:200px;
height:32pt;
white-space:pre;
display:block;
}
#input {
background-color:red;
}
#output {
background-color:blue;
}
#parse {
text-align:center;
font:24pt tahoma;
background-color:green;
}
</style>
<body onkeydown='span();'>
<span id='input' ></span>
<span id='output' ></span>
<span id='parse' onclick='parse();'>parse</span>
</body>
<script>
mult=new Array;
mult[0]=[0,1,2,3,4];
mult[1]=[1,0,3,4,2];
mult[2]=[2,4,0,1,3];
mult[3]=[3,2,4,0,1];
mult[4]=[4,3,1,2,0];
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);
inp+=chr;
input.textContent=inp;
}
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++;
}
out='';
c0=0;c4=0;
for (a=0;a<5;a++)
for (b=0;b<5;b++)
for (c=0;c<5;c++) {
x=mult[mult[a][b]][c];
out+='('+a+''+b+')'+c+'='+x+'<br>';
x=mult[b][mult[c][a]];
out+=b+'('+c+''+a+')='+x+'<br>';
}
output.innerHTML=out;
/* original
out='';
c0=0;c4=0;
for (a=0;a<5;a++)
for (b=0;b<5;b++)
for (c=0;c<5;c++) {
x=mult[mult[a][b]][c];
out+='('+a+''+b+')'+c+'='+x+'<br>';
if (x==0) c0++;
if (x==4) c4++;
x=mult[a][mult[b][c]];
out+=a+'('+b+''+c+')='+x+'<br>';
}
output.innerHTML=out;
*/
</script>
</html>
answered Apr 7 '15 at 13:50
JonMark PerryJonMark Perry
11.3k92438
$endgroup$
$begingroup$
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
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.
$endgroup$
– JonMark Perry
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
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
Apr 8 '15 at 13:01
add a comment |
$begingroup$
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 $yne z$ 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>
<html>
<title>Quasigroup Explorer</title>
<style>
span {
width:200px;
height:32pt;
white-space:pre;
display:block;
}
#input {
background-color:red;
}
#output {
background-color:blue;
}
#parse {
text-align:center;
font:24pt tahoma;
background-color:green;
}
</style>
<body onkeydown='span();'>
<span id='input' ></span>
<span id='output' ></span>
<span id='parse' onclick='parse();'>parse</span>
</body>
<script>
mult=new Array;
mult[0]=[0,1,2,3,4];
mult[1]=[1,0,3,4,2];
mult[2]=[2,4,0,1,3];
mult[3]=[3,2,4,0,1];
mult[4]=[4,3,1,2,0];
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);
inp+=chr;
input.textContent=inp;
}
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++;
}
out='';
c0=0;c4=0;
for (a=0;a<5;a++)
for (b=0;b<5;b++)
for (c=0;c<5;c++) {
x=mult[mult[a][b]][c];
out+='('+a+''+b+')'+c+'='+x+'<br>';
x=mult[b][mult[c][a]];
out+=b+'('+c+''+a+')='+x+'<br>';
}
output.innerHTML=out;
/* original
out='';
c0=0;c4=0;
for (a=0;a<5;a++)
for (b=0;b<5;b++)
for (c=0;c<5;c++) {
x=mult[mult[a][b]][c];
out+='('+a+''+b+')'+c+'='+x+'<br>';
if (x==0) c0++;
if (x==4) c4++;
x=mult[a][mult[b][c]];
out+=a+'('+b+''+c+')='+x+'<br>';
}
output.innerHTML=out;
*/
</script>
</html>
answered Apr 7 '15 at 13:50
JonMark PerryJonMark Perry
11.3k92438
$endgroup$
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 $yne z$ 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>
<html>
<title>Quasigroup Explorer</title>
<style>
span {
width:200px;
height:32pt;
white-space:pre;
display:block;
}
#input {
background-color:red;
}
#output {
background-color:blue;
}
#parse {
text-align:center;
font:24pt tahoma;
background-color:green;
}
</style>
<body onkeydown='span();'>
<span id='input' ></span>
<span id='output' ></span>
<span id='parse' onclick='parse();'>parse</span>
</body>
<script>
mult=new Array;
mult[0]=[0,1,2,3,4];
mult[1]=[1,0,3,4,2];
mult[2]=[2,4,0,1,3];
mult[3]=[3,2,4,0,1];
mult[4]=[4,3,1,2,0];
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);
inp+=chr;
input.textContent=inp;
}
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++;
}
out='';
c0=0;c4=0;
for (a=0;a<5;a++)
for (b=0;b<5;b++)
for (c=0;c<5;c++) {
x=mult[mult[a][b]][c];
out+='('+a+''+b+')'+c+'='+x+'<br>';
x=mult[b][mult[c][a]];
out+=b+'('+c+''+a+')='+x+'<br>';
}
output.innerHTML=out;
/* original
out='';
c0=0;c4=0;
for (a=0;a<5;a++)
for (b=0;b<5;b++)
for (c=0;c<5;c++) {
x=mult[mult[a][b]][c];
out+='('+a+''+b+')'+c+'='+x+'<br>';
if (x==0) c0++;
if (x==4) c4++;
x=mult[a][mult[b][c]];
out+=a+'('+b+''+c+')='+x+'<br>';
}
output.innerHTML=out;
*/
</script>
</html>
answered Apr 7 '15 at 13:50
JonMark PerryJonMark Perry
11.3k92438
edited Apr 8 '15 at 5:37
answered Apr 7 '15 at 13:50
JonMark PerryJonMark Perry
11.3k92438
answered Apr 7 '15 at 13:50
JonMark PerryJonMark Perry
11.3k92438
answered Apr 7 '15 at 13:50
JonMark PerryJonMark Perry
11.3k92438
11.3k92438
$begingroup$
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
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.
$endgroup$
– JonMark Perry
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
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
Apr 8 '15 at 13:01
add a comment |
$begingroup$
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
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.
$endgroup$
– JonMark Perry
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
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
Apr 8 '15 at 13:01
$begingroup$
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
Apr 8 '15 at 12:57
$begingroup$
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
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.
$endgroup$
– JonMark Perry
Apr 8 '15 at 12:59
$begingroup$
@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.
$endgroup$
– JonMark Perry
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
Apr 8 '15 at 13:00
$begingroup$
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
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
Apr 8 '15 at 13:01
$begingroup$
Cheers, Jon, I look forward to seeing whatever you come up with, even if it's negative!
$endgroup$
– Travis
Apr 8 '15 at 13:01
add a comment |
$begingroup$
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 $xlhd y = xyx^{-1}$ where the implied operations on the right hand side are the usual multiplication of permutations, that is, our operation now conjugates elements.
Let
$arightarrow(1,2,3)$
$brightarrow(1,4,2)$
$crightarrow(2,4,3)$
$drightarrow(1,3,4)$
and we get our quasigroup!
For example $alhd b=(1,2,3)lhd(1,4,2) = (1,2,3)(1,4,2)(1,2,3)^{-1}=(1,2,3)(1,4,2)(1,3,2)=(2,4,3)=c$
See also, conjugation and Quandles.
answered Dec 28 '18 at 17:04
Josh B.Josh B.
2,55511425
$endgroup$
$begingroup$
This is a nice description!
$endgroup$
– Travis
Dec 28 '18 at 18:36
add a comment |
$begingroup$
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 $xlhd y = xyx^{-1}$ where the implied operations on the right hand side are the usual multiplication of permutations, that is, our operation now conjugates elements.
Let
$arightarrow(1,2,3)$
$brightarrow(1,4,2)$
$crightarrow(2,4,3)$
$drightarrow(1,3,4)$
and we get our quasigroup!
For example $alhd b=(1,2,3)lhd(1,4,2) = (1,2,3)(1,4,2)(1,2,3)^{-1}=(1,2,3)(1,4,2)(1,3,2)=(2,4,3)=c$
See also, conjugation and Quandles.
answered Dec 28 '18 at 17:04
Josh B.Josh B.
2,55511425
$endgroup$
$begingroup$
This is a nice description!
$endgroup$
– Travis
Dec 28 '18 at 18:36
add a comment |
$begingroup$
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 $xlhd y = xyx^{-1}$ where the implied operations on the right hand side are the usual multiplication of permutations, that is, our operation now conjugates elements.
Let
$arightarrow(1,2,3)$
$brightarrow(1,4,2)$
$crightarrow(2,4,3)$
$drightarrow(1,3,4)$
and we get our quasigroup!
For example $alhd b=(1,2,3)lhd(1,4,2) = (1,2,3)(1,4,2)(1,2,3)^{-1}=(1,2,3)(1,4,2)(1,3,2)=(2,4,3)=c$
See also, conjugation and Quandles.
answered Dec 28 '18 at 17:04
Josh B.Josh B.
2,55511425
$endgroup$
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 $xlhd y = xyx^{-1}$ where the implied operations on the right hand side are the usual multiplication of permutations, that is, our operation now conjugates elements.
Let
$arightarrow(1,2,3)$
$brightarrow(1,4,2)$
$crightarrow(2,4,3)$
$drightarrow(1,3,4)$
and we get our quasigroup!
For example $alhd b=(1,2,3)lhd(1,4,2) = (1,2,3)(1,4,2)(1,2,3)^{-1}=(1,2,3)(1,4,2)(1,3,2)=(2,4,3)=c$
See also, conjugation and Quandles.
answered Dec 28 '18 at 17:04
Josh B.Josh B.
2,55511425
answered Dec 28 '18 at 17:04
Josh B.Josh B.
2,55511425
answered Dec 28 '18 at 17:04
Josh B.Josh B.
2,55511425
answered Dec 28 '18 at 17:04
Josh B.Josh B.
2,55511425
2,55511425
$begingroup$
This is a nice description!
$endgroup$
– Travis
Dec 28 '18 at 18:36
add a comment |
$begingroup$
This is a nice description!
$endgroup$
– Travis
Dec 28 '18 at 18:36
$begingroup$
This is a nice description!
$endgroup$
– Travis
Dec 28 '18 at 18:36
$begingroup$
This is a nice description!
$endgroup$
– Travis
Dec 28 '18 at 18:36
add a comment |
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$begingroup$
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