4
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.