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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.

Rodrigo de Azevedo
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Dan Glinski
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    Does a soduku have a unit element? – Mariano Suárez-Álvarez Aug 27 '16 at 16:27
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    Of course there is a way to tell from the table if the operation is associative: the table tells you everything there is to know about the multipliation. Of course, it is a somewhat laborious process! – Mariano Suárez-Álvarez Aug 27 '16 at 16:27
  • @Mariano Suárez-Álvarez if by that you mean an identity element then yes, though it may change. – Dan Glinski Aug 27 '16 at 16:31
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    As you saying that any sudoku table has an identity element? – Mariano Suárez-Álvarez Aug 27 '16 at 16:32
  • Maybe it would be easier to find the answer the other way around : take all groups of order $9$, construct their Cailey tables and see if they are Sudoku... – Arnaud D. Aug 27 '16 at 16:33
  • @ArnaudD., while that is easier, that will *not* answer the question... – Mariano Suárez-Álvarez Aug 27 '16 at 16:34
  • @Mariano Suárez-Álvarez, no I'm not saying that it must have an identity element because I don't know if Sudoku solutions are multiplication tables for groups and the lack of an identity element might be what is broken. I'm saying we have to account for the identity element changing when comparing two different solutions. – Dan Glinski Aug 27 '16 at 16:39
  • In a group, if $ab=1$, then $ba=1$. So there is an element in a Cayley table, corresponding to the neutral element, whose distribution is symmetric inside the table. Any sudoku falling that criterion (such as this one : https://upload.wikimedia.org/wikipedia/commons/thumb/3/31/Sudoku-by-L2G-20050714_solution.svg/2000px-Sudoku-by-L2G-20050714_solution.svg.png) would not be a Cayley table. – Joel Cohen Aug 27 '16 at 16:40
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    @DanGlinski, I suggest you pick any example of a Sudoky puzzle and check if it does have an identity element or not. – Mariano Suárez-Álvarez Aug 27 '16 at 16:42
  • I have checked 3 solutions a while ago and found one in each. It is a bit laborious and I would like to settle weather it is a group or not. One reason why it's hard to check is because you don't know what you're multiplying together when looking at the table. – Dan Glinski Aug 27 '16 at 16:44
  • It is quite difficult to find examples of that i nature, actually! can you share with us one of those examples? – Mariano Suárez-Álvarez Aug 27 '16 at 16:50
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    That is somewhat irrrelevant, really. If there is an identity element, for example, one of the columns has to be equal to the row symmetric to it. That rarely happens. – Mariano Suárez-Álvarez Aug 27 '16 at 16:51
  • I'm sorry, it was about 3 months ago when I did that and I can't find which ones they were on google images. – Dan Glinski Aug 27 '16 at 16:53
  • Well, google for one and try again. :-) – Mariano Suárez-Álvarez Aug 27 '16 at 16:54
  • I looked at the table for this one and couldn't find an identity element...of course I could have made a mistake. #64 http://www.puzzles.ca/sudoku_puzzles/sudoku_easy_063_solution.html – Dan Glinski Aug 27 '16 at 17:37
  • There are only two groups of order $9$: $\mathbb{Z}_9$ and $\mathbb{Z}_3^2$. It's easy to check whether a sudoku is the multiplication table for either group. – anomaly Aug 27 '16 at 23:36
  • @anomaly, How do you check? – Dan Glinski Aug 28 '16 at 00:20
  • @DanGlinski: Just check whether there's a permutation of $\{1, \dots, 9\}$ in the suduko that reproduces either table. – anomaly Aug 28 '16 at 00:28
  • @anomaly: It's not that simple, because it seems to me that Dan allows the possibility that the ordering of the rows need not match with the ordering of columns. Even so, it is still easy to construct sudoku's that don't come from a reordered Cayley table. – Jyrki Lahtonen Aug 29 '16 at 19:31

3 Answers3

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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

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

Joel Cohen
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    In fact, the failure is more dramatic: since every group of order 9 is abelian the whole multipication table is symmetric. – Mariano Suárez-Álvarez Aug 27 '16 at 17:36
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    But what if you allow swapping the rows and columns around? Wouldn't that destroy the symmetry in the table? – Dan Glinski Aug 27 '16 at 17:40
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    @DanGlinski, if you intend to permute rows and columns *independently* yes, but no one understood that in your question, as interpreting a table as a multiplication table sort of goes against that. – Mariano Suárez-Álvarez Aug 27 '16 at 17:57
  • I'm sorry I wasn't clear. What would you call the table? It does say what happens when you combine elements in the table, they're just not in any particular order. – Dan Glinski Aug 27 '16 at 18:31
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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.
Jyrki Lahtonen
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Another approach: We are considering a group of order $9 = 3^2$ so a prime squared. It is well known that groups of order $p^2$ must be abelian. (abelian = commutative) Now a soduku can never represent an abelian group, as it is never symmetric. (Symmetric as in matrices, that means symmetric wrt. the diagonal.)

flawr
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