Some ternary quasigroups over small sets.
 
Page one contains sections 1 through 4.
Page two contains sections 5 through 10.
Page three contains sections 11 and 12.
Page four contains sections 13 through 15.
This is page five, with sections 16 through 18.


§16A Commutative complementation. This is best explained with examples. Define names for six particular operations:

O= 4:34877
N= 4:35635
M= 4:32473
 
L= 4:29035
K= 4:33549
J= 4:28801

None of those six operations has any commutativity in its own right, but each has other operations as commutative complements as listed below. These six operations form a closed set under commutative complementation.

commutative complements of O, N, M, L, K, J
sinisterior: O (p, q, r) = N (q, p, r) exterior: O (p, q, r) = M (r, q, p) dexterior: O (p, q, r) = L (p, r, q)
M (p, q, r) = K (q, p, r) N (p, q, r) = J (r, q, p) N (p, q, r) = K (p, r, q)
L (p, q, r) = J (q, p, r) L (p, q, r) = K (r, q, p) M (p, q, r) = J (p, r, q)
r-cyclical: O (p, q, r) = J (r, p, q) q-cyclical: O (p, q, r) = K (q, r, p)
N (p, q, r) = M (r, p, q) N (p, q, r) = L (q, r, p)
M (p, q, r) = L (r, p, q) M (p, q, r) = N (q, r, p)
L (p, q, r) = N (r, p, q) L (p, q, r) = M (q, r, p)
K (p, q, r) = O (r, p, q) K (p, q, r) = J (q, r, p)
J (p, q, r) = K (r, p, q) J (p, q, r) = O (q, r, p)

To set up another example, define names for three more operations:

X = 4:35451  Y = 4:35465  Z = 4:36041

Each of the them is commutative (with itself) in exactly one way: X sinisterior, Y exterior, and Z dexterior. They too form a closed set under commutative complementation:

commutative complements of X, Y, Z
sinisterior: X (p, q, r) = X (q, p, r) exterior: Y (p, q, r) = Y (r, q, p) dexterior: Z (p, q, r) = Z (p, r, q)
Y (p, q, r) = Z (q, p, r) Z (p, q, r) = X (r, q, p) X (p, q, r) = Y (p, r, q)
r-cyclical: X (p, q, r) = Z (r, p, q) q-cyclical: X (p, q, r) = Y (q, r, p)
Y (p, q, r) = X (r, p, q) Y (p, q, r) = Z (q, r, p)
Z (p, q, r) = Y (r, p, q) Z (p, q, r) = X (q, r, p)


§16B. One might hope for associative complementation, but the situation becomes murkier. Define these:

F = 3::1  G = 3::2  H = 3::5

It turns out that F has more than one associative complement:

F (F (p, q), r) = G (p, G (q, r)) = H (p, H (q, r))

Also, the distinction between operations G and H is lost when each is associated on the right. Meanwhile, 3::4 has no associative complement, and 3::0 has only itself.


§17 Disjunction. Two ternary operations O and N are disjoint if O (p, q, r) ≠ N (p, q, r) for all values of p, q, r. When C = n, there might exist a set of n operations that are pairwise disjoint; but never a set of n + 1 elements.

A similar idea of course applies to binary operations. If two unary operations are disjoint, one is said to be a derangement of the other.

The 24 ternary operations with C = 3 happen to form 8 sets of 3 pairwise disjoint operations, listed below with their conjugation classes.

3:0 – A
3:15 – A
3:16 – A
3:3 – B
3:12 – B
3:19 – B
3:5 – C
3:10 – C
3:21 – C
3:6 – D
3:9 – D
3:22 – D
3:7 – E
3:8 – E
3:23 – E
3:4 – K
3:11 – F
3:20 – F
3:2 – J
3:13 – G
3:18 – G
3:1 – I
3:14 – H
3:17 – H

Here is a example elaborated for direct comparison:

3:4, 3:11, 3:20
aaa =
a, b, c
aab =
c, a, b
aac =
b, c, a
aba =
b, c, a
abb =
a, b, c
abc =
c, a, b
aca =
c, a, b
acb =
b, c, a
acc =
a, b, c
baa =
b, c, a
bab =
a, b, c
bac =
c, a, b
bba =
c, a, b
bbb =
b, c, a
bbc =
a, b, c
bca =
a, b, c
bcb =
c, a, b
bcc =
b, c, a
caa =
c, a, b
cab =
b, c, a
cac =
a, b, c
cba =
a, b, c
cbb =
c, a, b
cbc =
b, c, a
cca =
b, c, a
ccb =
a, b, c
ccc =
c, a, b

Because the outputs of the three operations within each set happen to be related by the permutation abca, disjunction might be confused with conjugation. Recall however that in conjugation, all inputs and outputs are permuted. By contrast, the three operations in each set above are related by permutation of not their inputs, but only their outputs. This is why operations that reside in different conjugation classes might appear within the same set here.

When C = 4, there do exist sets of four ternary operations that are pairwise disjoint. In the example below, the outputs are not related by permutation.

4:0, 4:16520, 4:38775, 4:55295
aaa =
a, b, c, d
aab =
b, a, d, c
aac =
c, d, a, b
aad =
d, c, b, a
aba =
b, a, d, c
abb =
a, b, c, d
abc =
d, c, b, a
abd =
c, d, a, b
aca =
c, d, a, b
acb =
d, c, b, a
acc =
a, b, c, d
acd =
b, a, d, c
ada =
d, c, b, a
adb =
c, d, a, b
adc =
b, a, d, c
add =
a, b, c, d
baa =
b, d, a, c
bab =
a, b, c, d
bac =
d, c, b, a
bad =
c, a, d, b
bba =
a, c, b, d
bbb =
b, a, d, c
bbc =
c, d, a, b
bbd =
d, b, c, a
bca =
d, b, c, a
bcb =
c, d, a, b
bcc =
b, a, d, c
bcd =
a, c, b, d
bda =
c, a, d, b
bdb =
d, c, b, a
bdc =
a, b, c, d
bdd =
b, d, a, c
caa =
c, a, d, b
cab =
d, c, b, a
cac =
a, b, c, d
cad =
b, d, a, c
cba =
d, b, c, a
cbb =
c, d, a, b
cbc =
b, a, d, c
cbd =
a, c, b, d
cca =
a, c, b, d
ccb =
b, a, d, c
ccc =
c, d, a, b
ccd =
d, b, c, a
cda =
b, d, a, c
cdb =
a, b, c, d
cdc =
d, c, b, a
cdd =
c, a, d, b
daa =
d, c, b, a
dab =
c, d, a, b
dac =
b, a, d, c
dad =
a, b, c, d
dba =
c, d, a, b
dbb =
d, c, b, a
dbc =
a, b, c, d
dbd =
b, a, d, c
dca =
b, a, d, c
dcb =
a, b, c, d
dcc =
d, c, b, a
dcd =
c, d, a, b
dda =
a, b, c, d
ddb =
b, a, d, c
ddc =
c, d, a, b
ddd =
d, c, b, a

The outputs of these four operations do not form all 24 = 4! possible sequences. Specifically:


§18 Steiner systems. Well known is that a Steiner system can be used to construct an idempotent commutative quasigroup.

§18A. The Steiner triple system S (2, 3, 9) will produce a binary quasigroup. To find one (there are many), a procedure starts with this set:

T = {a, b, c, d, e, f, g, h, i}

Of the 84 size-3 subsets of T, 12 have been selected to appear in the 3-set column. Meanwhile, in the 2-sets column, each of the 36 size-2 subsets of T appears exactly once. Within each row, each 2-set is a subset of the 3-set.

3-set2-sets
#1{a, b, c}{a, b}, {a, c}, {b, c}
#2{a, d, e}{a, d}, {a, e}, {d, e}
#3{a, f, g}{a, f}, {a, g}, {f, g}
#4{a, h, i}{a, h}, {a, i}, {h, i}
#5{b, d, f}{b, d}, {b, f}, {d, f}
#6{b, e, h}{b, e}, {b, h}, {e, h}
#7{b, g, i}{b, g}, {b, i}, {g, i}
#8{c, d, i}{c, d}, {c, i}, {d, i}
#9{c, e, g}{c, e}, {c, g}, {e, g}
#10{c, f, h}{c, f}, {c, h}, {f, h}
#11{d, g, h}{d, g}, {d, h}, {g, h}
#12{e, f, i}{e, f}, {e, i}, {f, i}

To define operation O (p, q), also written Opq:

Here is the Cayley table for O:

aa=aab=cac=b ad=eae=daf=g ag=fah=iai=h
ba=cbb=bbc=a bd=fbe=hbf=d bg=ibh=ebi=g
ca=bcb=acc=c cd=ice=gcf=h cg=ech=fci=d
da=edb=fdc=i dd=dde=adf=b dg=hdh=gdi=c
ea=deb=hec=g ed=aee=eef=i eg=ceh=bei=f
fa=gfb=dfc=h fd=bfe=iff=f fg=afh=cfi=e
ga=fgb=igc=e gd=hge=cgf=a gg=ggh=dgi=b
ha=ihb=ehc=f hd=ghe=bhf=c hg=dhh=hhi=a
ia=hib=gic=d id=cie=fif=e ig=bih=aii=i

The 3-sets can be partitioned so that each element of T appears exactly once within each partition, as below. This is an example of a Kirkman triple system.

{a, b, c}, {d, g, h}, {e, f, i}
{a, d, e}, {b, g, i}, {c, f, h}
{a, f, g}, {b, e, h}, {c, d, i}
{a, h, i}, {b, d, f}, {c, e, g}


§18B. Similar is the next example, which uses a Steiner quadruple system S (3, 4, 8) to produce a ternary quasigroup. It starts with this set:

U = {a, b, c, d, e, f, g, h}

Of the 70 size-4 subsets of U, 14 have been selected to appear in the 4-set column. Meanwhile, in the 3-sets column, each of the 56 size-3 subsets of U appears exactly once. Within each row, each 3-set is a subset of the 4-set.

4-set3-sets
#1{a, b, c, d}{a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}
#2{a, b, e, f}{a, b, e}, {a, b, f}, {a, e, f}, {b, e, f}
#3{a, b, g, h}{a, b, g}, {a, b, h}, {a, g, h}, {b, g, h}
#4{a, c, e, g}{a, c, e}, {a, c, g}, {a, e, g}, {c, e, g}
#5{a, c, f, h}{a, c, f}, {a, c, h}, {a, f, h}, {c, f, h}
#6{a, d, e, h}{a, d, e}, {a, d, h}, {a, e, h}, {d, e, h}
#7{a, d, f, g}{a, d, f}, {a, d, g}, {a, f, g}, {d, f, g}
#8{b, c, e, h}{b, c, e}, {b, c, h}, {b, e, h}, {c, e, h}
#9{b, c, f, g}{b, c, f}, {b, c, g}, {b, f, g}, {c, f, g}
#10{b, d, e, g}{b, d, e}, {b, d, g}, {b, e, g}, {d, e, g}
#11{b, d, f, h}{b, d, f}, {b, d, h}, {b, f, h}, {d, f, h}
#12{c, d, e, f}{c, d, e}, {c, d, f}, {c, e, f}, {d, e, f}
#13{c, d, g, h}{c, d, g}, {c, d, h}, {c, g, h}, {d, g, h}
#14{e, f, g, h}{e, f, g}, {e, f, h}, {e, g, h}, {f, g, h}

To define operation N (p, q, r), also written Npqr:

On another page is displayed the lengthy Cayley table in full. It turns out that this operation is also fully associative. However, other Steiner systems might lead to non-associative operations.

Here is a partitioning of the 4-sets comparable to the Kirkman above:

{a, b, c, d}, {e, f, g, h}
{a, b, e, f}, {c, d, g, h}
{a, b, g, h}, {c, d, e, f}
{a, c, e, g}, {b, d, f, h}
{a, c, f, h}, {b, d, e, g}
{a, d, e, h}, {b, c, f, g}
{a, d, f, g}, {b, c, e, h}

§18C. The two examples so far have been of the form (t, t + 1, n). A system of the form (t, t + 2, n) can be considerably more complicated.


Page one contains sections 1 through 4.
Page two contains sections 5 through 10.
Page three contains sections 11 and 12.
Page four contains sections 13 through 15.
This is page five, with sections 16 through 18.

directory of files
numbered pagesoperation charts

page one (§1 – §4):

page two (§5 – §10):

page three (§11 – §12)

page four (§13 – §15)

page five (§16 – §18):

binary:

ternary: