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§ 1B A caution pertaining to Cayley tables. Below are two ordinary Latin squares, the same except that the rows have been rearranged. For most combinatorical purposes, the two squares would be regarded as essentially the same thing.
| Latin square 1
|
|---|
| a | c | d | b
| | d | b | a | c
| | b | d | c | a
| | c | a | b | d
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|
|
| Latin square 2
|
|---|
| c | a | b | d
| | a | c | d | b
| | d | b | a | c
| | b | d | c | a
|
|
Below, each of these Latin squares has been installed as the body of the Cayley table for a binary quasigroup:
F = 4::91
from
Latin square 1
| second input
|
|---|
| a | b | c | d
|
|---|
| first input | a
| a | c
| d | b
|
|---|
| b
| d | b
| a | c
|
|---|
| c
| b | d
| c | a
|
|---|
| d
| c | a
| b | d
|
|---|
|
|
G = 4::292
from
Latin square 2
| second input
|
|---|
| a | b | c | d
|
|---|
| first input | a
| c | a
| b | d
|
|---|
| b
| a | c
| d | b
|
|---|
| c
| d | b
| a | c
|
|---|
| d
| b | d
| c | a
|
|---|
|
For all p, observe that F (p, p) = p, while by contrast G (p, p) ≠ p. With the loss of idempotence, F and G would be deemed non-isomorphic for virtually all algebraic purposes, even though they are based on equivalent Latin squares. What happened? The outputs of the operations were changed, but not the inputs, resulting in an inconsistency.
The same principle naturally applies to Latin cubes and ternary quasigroups.