# Is this the Cayley table of a group? (Part 2)

In my previous post I posed the following question: suppose you are given a table for a binary operation such that

1. there is a two-sided identity,
2. every element has a two-sided inverse, and
3. 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 $i^2=j^2=k^2=1$. Then $(i*i)*j=j$ and $i*(i*j)=i*k=-j$. Thus the operation is not associative and we have an example of order 8. (I found this example here.)

$\begin{array}{|c||c|c|c|c|c|c|c|c|}\hline * & 1 & -1 & i & -i & j & -j & k & -k \\\hline\hline 1 & 1 & -1 & i & -i & j & -j & k & -k \\\hline -1 & -1 & 1 & -i & i & -j & j & -k & k \\\hline i & i & -i & 1 & -1 & k & -k & -j & j \\\hline -i & -i & i & -1 & 1 & -k & k & j & -j \\\hline j & j & -j & -k & k & 1 & -1 & i & -i \\\hline -j & -j & j & k & -k & -1 & 1 & -i & 0 \\\hline k & k & -k & j & -j & -i & i & 1 & -1 \\\hline -k & -k & k & -j & j & i & -i & -1 & 1 \\\hline \end{array}$

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.

$\begin{array}{|c||c|c|c|c|c|}\hline * & 1 & 2 & 3 & 4 & 5 \\\hline\hline 1 & 1 & 2 & 3 & 4 & 5 \\\hline 2 & 2 & 1 & 4 & 5 & 3 \\\hline 3 & 3 & 5 & 1 & 2 & 4 \\\hline 4 & 4 & 3 & 5 & 1 & 2 \\\hline 5 & 5 & 4 & 2 & 3 & 1 \\\hline \end{array}$

Indeed it satisfies (1)-(3), but it fails to be associative. For example, $4*(3*2)=4*5=2$ and $(4*3)*2=5*2=4$. (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.