# Characterizing group operation properties by its multiplication table

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.