In my previous post I posed the following question: suppose you are given a table for a binary operation such that
- there is a two-sided identity,
- every element has a two-sided inverse, and
- the table is a Latin square (that is, each symbol occurs exactly once in each row and each column).
Is it the Cayley table of a group? If so, prove it. If not, find a minimal counter-example.
The answer, not surprisingly, is no. It need not be a group.
A set satisfying properties (2) and (3) is called a quasigroup, and one satisfying (1), (2), and (3) is called a loop. Thus we would like to find the smallest loop that is not a group.
We can create a simple example of a loop that is not a group by modifying the quaternions. Keep all the usual products, but assume that . Then and . Thus the operation is not associative and we have an example of order 8. (I found this example here.)
What about a minimal example? It is trivial to check that any 1×1, 2×2, and 3×3 table that satisfies (1)-(4) is a group. The 4×4 case requires a few more cases to check, but likewise any 4×4 loop is a group.
Thus we turn to the 5×5 case. This is our minimal case. The paper “Cayley Tables and Associativity” (pdf), by R. P. Burn, (The Mathematical Gazette, Vol. 62, No. 422, (Dec., 1978), pp. 278-281) contains the following example.
Indeed it satisfies (1)-(3), but it fails to be associative. For example, and . (Actually, we can easily see that this fails to be a group by Lagrange’s Theorem: every element is its own inverse, and hence has order 2, but 2 does not divide 5.) This paper has an interesting discussion of tables that have certain types associative products.