How would one go about to proving that in a Cayley table, every elements appear once and only once in every row and in every column?
Consider group $(G,\cdot)$. Let $x^{-1}$ be the inverse of $x$, and let $e$ be the identity.
$x^{-1} x = e$
$x y = g = x z$
$y = e y = (x^{-1} x) y = x^{-1} (x y) = x^{-1}g = x^{-1} (x z) = (x^{-1} x) z = e z = z,$
which seems to be a contradiction.
Best Answer
You didn't derive a contradiction, you've actually just proved left cancellation: If $xy = xz$ then $y=z$, i.e., you may cancel the $x$'s off the left side.
This means that in a row, each element may only occur at most once, or else it would violate the left cancellation law. Similarly, there is right cancellation, which will imply the same thing for the columns.
What's left is to prove every element occurs at least once in each row/column. Consider the $x$ row, and some arbitrary element $y$. Can you find an element $z$ such that $xz=y$ (and for the column case, $zx=y$)? Hopefully this will be easy to spot!