I want to know if the popular Sudoku puzzle is a Cayley table for a group.
Methods I've looked at: Someone I've spoken to told me they're not because counting the number of puzzle solutions against the number of tables with certain permutations of elements, rows and columns, the solutions are bigger than the tables, but I can't see why because I don't know how to count the different tables for a group of order 9, and then permute the rows, columns and elements in different ways. Also I believe the rotations/reflections will matter in comparing these numbers too. It would also be nice if there was a way to know if the operation is associative just from the table.
Best Answer
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
$$ \begin{array}{r|lll|lll|lll} +&0&1&2&3&4&5&6&7&8\\ \hline 0&0&1&2&3&4&5&6&7&8\\ 3&3&4&5&6&7&8&0&1&2\\ 6&6&7&8&0&1&2&3&4&5\\ \hline 1&1&2&3&4&5&6&7&8&0\\ 4&4&5&6&7&8&0&1&2&3\\ 7&7&8&0&1&2&3&4&5&6\\ \hline 2&2&3&4&5&6&7&8&0&1\\ 5&5&6&7&8&0&1&2&3&4\\ 8&8&0&1&2&3&4&5&6&7\\ \end{array}$$