# [Maths-Education] Re: Maths-Education Digest, Vol 50, Issue 3

Ng Foo Keong lefouque at gmail.com
Fri Dec 5 21:25:14 GMT 2008

```Does a completed Soduko (9X9) grid embody a  group (of order 9)?
Consider the example
*
2 5 3 1 9 4 8 7 6
7 9 4 6 3 8 5 2 1
8 1 6 7 2 5 4 9 3
5 8 1 9 4 6 7 3 2
3 2 9 8 1 7 6 5 4
6 4 7 2 5 3 1 8 9
1 7 8 3 6 9 2 4 5
9 6 5 4 8 2 3 1 7
4 3 2 5 7 1 9 6 8

Maybe there is a shorter proof:-
Suppose there were a group G and consider its group action
on itself.  The permutation that transforms row 1 to row 2 is
p = (2 7)(5 9 3 4 8 5)(1 6)
Let g be the element of G such that p: x |---> g*x.
Note that p is of order 6, thus the cyclic subgroup <g> is of
order 6, which does not divide |G| = 9, contradicting Lagrange's
Theorem.                                               (Q.E.D.)
---------------------------------------------------------------
Does any Soduko grid represent a group?

Suppose we have the condition that the horizontal and vertical
headings of the group table are in the same order:-
Consider the locale (one of the nine 3×3 sub-grids that
should contain all digits 1 thru 9) that contains the identity e.
Take another element x in that locale in the same _row_ as e.
Since e * x = x, we know x appears as a column heading, and
correspondingly as a row heading for that locale.  But then
x * e = x, which appears in the locale in the same _column_ as e.
Thus x will be a repeat element in the locale, contradicting the
rules of Sudoku.

If we relax the above condition, then an easy solution would be:-
1 2 3 4 5 6 7 8 9
4 5 6 7 8 9 1 2 3
7 8 9 1 2 3 4 5 6
2 3 4 5 6 7 8 9 1
5 6 7 8 9 1 2 3 4
8 9 1 2 3 4 5 6 7
3 4 5 6 7 8 9 1 2
6 7 8 9 1 2 3 4 5
9 1 2 3 4 5 6 7 8

regards,
Ng, Foo Keong
Singapore
```

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