Characterizing group operation properties by its multiplication table

Question

Let $G = \{x_1,\dots, x_n\}$ be a set equipped with an operation $*$. Let $A = [a_{ij}]$ be its multiplication table, $a_{ij} = x_i*x_j$. Assume $G$ has a identity $e$ (such that $e*x=x*e=x$ for all $x\in G$). Show that every element $x\in G$ has a two sided inverse (i.e., there is a $x'\in G$ with $x*x' = x'*x = e$) if and only if the multiplication table $A$ is an Latin square; that is, no $x\in G$ is repeated in any row or column (= every row or column is a permutation of $G$).

If every $x$ has a inverse, then given $a_{ij} = a_{ik}$ for $1\le i,j,k\le n$ then $x_i*x_j = x_i*x_k$ and multiplying by $x_i^{-1}$ in the left on both sides we get $x_j = x_k$ then no two elements in distinct positions in line $i$ are equal. The same applies for columns using inverses in the right. Then $A$ is Latin square.

But I'm struggling with the converse. Given $x_i\in G$, there is a $a_{ij}$ in line $i$ such that $x_i*x_j = e$ and in the column $i$ there is a $a_{ki}$ such that $x_k*x_i = e$, and I must show that $x_j = x_k$. I tried other similar ways to write these multiplications but I don't see how to show the equality without using associativity (which is not assumed).

Thanks for the help.

Answer 1

Without associativity, you are right that the statement is false. In fact,

• the multiplication table of a loop is a Latin square, and
• the second multiplication table of this MSE post gives an example of a (non-associative) loop in which right and left inverses differ.