Is a Sudoku a Cayley table for a group?

A Cayley table for a group can never be a sudoku. Assume you have a $9 \times 9$ Cayley table for a group of order $9$, and say your identity element is at index $1 \le i \le 9$. Then row $i$ and column $i$ are symmetric to each other because they correspond to multiplication with the identity. In particular, if you look at the $3\times3$ sub-square containing the element of coordinates $(i,i)$, this square has duplicates (because it contains symmetric elements of row $i$ and column $i$). So the table isn't a Sudoku.

If you allow to swap the rows or columns, this is possible. Take the table of $G = \mathbb{Z}/9\mathbb{Z}$ (I wrote $0$ instead of $9$ for convenience): $$ \begin{array}{r|lllllllll} +&0&1&2&3&4&5&6&7&8\\ \hline 0&0&1&2&3&4&5&6&7&8\\ 1&1&2&3&4&5&6&7&8&0\\ 2&2&3&4&5&6&7&8&0&1\\ 3&3&4&5&6&7&8&0&1&2\\ 4&4&5&6&7&8&0&1&2&3\\ 5&5&6&7&8&0&1&2&3&4\\ 6&6&7&8&0&1&2&3&4&5\\ 7&7&8&0&1&2&3&4&5&6\\ 8&8&0&1&2&3&4&5&6&7\\ \end{array}$$

And swap the rows in order $0,3,6,1,4,7,2,5,8$, to obtain the Sudoku

$$ \begin{array}{r|lll|lll|lll} +&0&1&2&3&4&5&6&7&8\\ \hline 0&0&1&2&3&4&5&6&7&8\\ 3&3&4&5&6&7&8&0&1&2\\ 6&6&7&8&0&1&2&3&4&5\\ \hline 1&1&2&3&4&5&6&7&8&0\\ 4&4&5&6&7&8&0&1&2&3\\ 7&7&8&0&1&2&3&4&5&6\\ \hline 2&2&3&4&5&6&7&8&0&1\\ 5&5&6&7&8&0&1&2&3&4\\ 8&8&0&1&2&3&4&5&6&7\\ \end{array}$$

It is easy to construct a Sudoku that cannot be a Cayley table. Not even if you label the rows and columns independently from each other. Consider the following. $$ \begin{array}{ccc|ccc|ccc} 1&2&3&4&5&6&7&8&9\\ 4&5&6&7&8&9&1&3&2\\ 7&8&9&1&2&3&4&5&6\\ \hline 2&3&1&5&6&4&9&7&8\\ 5&6&4&8&9&7&2&1&3\\ 8&9&7&2&3&1&6&4&5\\ \hline 3&1&2&6&4&5&8&9&7\\ 6&4&5&9&7&8&3&2&1\\ 9&7&8&3&1&2&5&6&4 \end{array} $$ My argument only needs the two first rows. The rest were filled in just to leave no doubt about the fact that those two rows form a part of a complete Sudoku.

We see that we get the second row from the first by applying the permutation $\sigma=(147)(258369)$ to it. This is all we need to prove that this is not a Cayley table. If in a Cayley table the first row has $a$ as the common left factor and the second row has $b$ as the common left factor, we get the second row from the first by multiplying everything from the left by $c:=ba^{-1}$. If $n=\operatorname{ord}(c)$ then the permutation $\sigma$ bringing the first row to the second will then only have cycles of length $n$ as the cycles act transitively on right cosets $Hx$, $H=\langle c\rangle, x\in G$.

But our $\sigma$ has cycle type $(3,6)$. This already gives two obvious contradictions:

• the lengths of the cycles vary, and
• $\sigma$ has order six, so cannot be an element of a group of order nine.