4
$\begingroup$

A group must have a Cayley table in which each row and column has one and only one of each element. This can be proved by considering the opposite: suppose one row of a set’s Cayley table did not contain a particular element. That is, let AX not equal B, whatever X is. Let C be A’s inverse, such that CA = I. We easily reach a contradiction of associativity with ACB. (A*C)B = IB = B. On the other hand, A*(CB) = AX, does not equal B. Hence, each row and column of a group's Cayley table must contain exactly one of each element.

Two questions then arise:

Does a Cayley table in which each row and column has exactly one of each element guarantee that it is a group?

For a given number of elements, how many possible non-isomorphic groups are there? For instance, there is only one group with three elements. I think there are 4 groups with four elements.

$\endgroup$
  • 3
    $\begingroup$ Google for "quasi-group" or "latin square". $\endgroup$ – Mariano Suárez-Álvarez Jul 5 '17 at 4:33
  • $\begingroup$ Most tables that satisfy the sudoku conditions are not group tables. For locally handled examples see here or here. The last question is too broad. Books have been filled to give (partial) answers. And, there are only two groups with four elements up to isomorphism. $\endgroup$ – Jyrki Lahtonen Jul 5 '17 at 5:48
  • 1
    $\begingroup$ @JyrkiLahtonen Also math.stackexchange.com/questions/309205/… $\endgroup$ – Tobias Kildetoft Jul 5 '17 at 7:14
  • $\begingroup$ Thanks @TobiasKildetoft! I had somehow missed that thread. That goes a bit deeper! $\endgroup$ – Jyrki Lahtonen Jul 5 '17 at 8:16
  • $\begingroup$ @JyrkiLahtonen Right, though it is in some sense from the opposite direction, as it starts with the group (also, I never got around to checking what happens for some of the small groups of square order that do not satisfy the condition given in the answer there. Not that I have a good idea for how to do so easily). $\endgroup$ – Tobias Kildetoft Jul 5 '17 at 8:23
4
$\begingroup$

No! Such a Cayley table will not necessarily satisfy the associative law, which is an absolutely critical property in group theory.

You can easily construct a counterexample on 5 elements.

a b c d e
b c d e a 
c a e b d
d e b a c
e d a c b

Every symbol occurs exactly once in every row or column, and $a$ is the identity element, but this is not associative: $(b*b)*b = a$ while $b*(b*b) = d$. (The multiplication $xy$ is defined by looking up row $x$ and column $y$.)

In general, Cayley tables are not a good way of constructing groups. Even though Cayley tables are usually presented in beginner texts due to their resemblance to the familiar addition/multiplication tables, the associative property is difficult to check in a Cayley table. Instead, groups are usually constructed in terms of some operation which is already known to be associative, such as function composition, integer addition or multiplication, or (once you get off the ground) previously constructed groups.

$\endgroup$

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.