Structural constraints imposed by associativity on group multiplication table

Question

The closure property requires that all elements in the multiplication table belong to the group. The existence of identity requires the identity row and column (where the other operand is left unchanged). The existence of a unique inverse for every element requires that the identity element appears in every row and column exactly once. What constraint does the associativity property impose on the table?

The rearrangement theorem states that each row and column in the group multiplication table lists each group element exactly once. How do we show that a table that satisfies rearrangement theorem conditions and the existence of an identity element also implies associativity?

Answer 1

Rearrangement condition means multiplication table is Latin square, and if enumerate columns from left to right and rows from top to bottom and require identity to be denoted by $0$, then existence of identity is equal to the square been reduced.

The smallest example which is not group is $$\begin{array}{c|ccccc} & 0 & 1 & 2 & 3 & 4 \\\hline 0 & 0 & 1 & 2 & 3 & 4 \\ 1 & 1 & 0 & 3 & 4 & 2 \\ 2 & 2 & 3 & 4 & 0 & 1 \\ 3 & 3 & 4 & 1 & 2 & 0 \\ 4 & 4 & 2 & 0 & 1 & 3 \end{array}$$

It's not associative, as $2 \circ (3 \circ 4) = 2 \circ 0 = 2 \neq 4 = 0 \circ 4 = (2 \circ 3) \circ 4$.

Answer 2

A structural constraint from the associative law is the following: For every $i,j,k\in\{1,\dots,n\}$: $$i*j=k\Longrightarrow \sigma_i\sigma_j=\sigma_k\tag1$$ where $\sigma_i\in S_n$ is defined by: $$\sigma_i(j):=i*j\tag2$$

For example, for the following Latin square: $$\begin{array}{c|ccccc} * & 1 & 2 & 3 & 4 & 5 \\\hline 1 & 1 & 2 & 3 & 4 & 5 \\ 2 & 2 & 1 & 4 & 5 & 3 \\ 3 & 3 & 4 & 5 & 1 & 2 \\ 4 & 4 & 5 & 2 & 3 & 1 \\ 5 & 5 & 3 & 1 & 2 & 4 \end{array}$$ $(2)$ yields: \begin{alignat}{1} &\sigma_1=() \\ &\sigma_2=(12)(345) \\ &\sigma_3=(13524) \\ &\sigma_4=(14325) \\ &\sigma_5=(15423) \\ \end{alignat} Now, $3*5=2$ but $\sigma_3\sigma_5=(125)\ne\sigma_2$. According to $(1)$, this is sufficient to state that this composition law is not associative.

Answer 3

Given a set $X$ equipped with a binary operation $\ast : X \times X \to X$, the property that each element of $X$ occurs exactly once in the Cayley table (multiplication table) for $\ast$ is called the Latin square property, and if $\ast$ satisfies the Latin square property we call the structure $(X, \ast)$ a quasigroup. A quasigroup $(X, \ast)$ equipped with an identity is called a loop, and a group is precisely a quasigroup whose operation is associative. In this language the question asks how to show that a loop is associative---which is not true.

A particularly interesting example of a nonassociative loop is the set of nonzero octonions equipped with octonion multiplication. The loops of order $\leq 4$ are all groups, but up to isomorphism there are $6$ loops with exactly $5$ elements, and only $1$ is a group. The $5$ nonassociative loops of order $5$ are specified by the following Cayley tables, recorded here as a $5 \times 5$ matrices. \begin{multline} \left[\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 1 & 4 & 5 & 3 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 2 & 3 & 1 \\ 5 & 3 & 1 & 2 & 4 \end{array}\right] , \left[\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 1 & 4 & 5 & 3 \\ 3 & 4 & 5 & 2 & 1 \\ 4 & 5 & 1 & 3 & 2 \\ 5 & 3 & 2 & 1 & 4 \end{array}\right] , \left[\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 1 & 4 & 5 & 3 \\ 3 & 5 & 1 & 2 & 4 \\ 4 & 3 & 5 & 1 & 2 \\ 5 & 4 & 2 & 3 & 1 \end{array}\right] , \\ \left[\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 1 & 4 & 5 & 3 \\ 3 & 5 & 2 & 1 & 4 \\ 4 & 3 & 5 & 2 & 1 \\ 5 & 4 & 1 & 3 & 2 \end{array}\right] , \left[\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 1 & 5 & 4 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 2 & 3 & 1 \\ 5 & 1 & 4 & 2 & 3 \end{array}\right]. \end{multline} The examples in the two previously posted answers are isomorphic to the first. The third table defines a loop in which every element squares to the identity, which in particular implies power-alternativity (in fact the operation is even flexible); the other four are not power-alternative. This question asks for a concrete interpretation of that loop.