Rich wrote:
<<
. . . trying out non-commutative squares. This would let us
return to the idea of using all the members of a well-defined set.
The two most interesting cases to try seem to be using one of
the non-commutative groups with 16 elements to make a 4x4, or
using 2x2 matrices with elements 0,1 or maybe +-1, also to
make a 4x4 square. The 0,1 matrix case would need a magic
constant of [00,00]. You need some rules on the order of
combining things, and whether the up-going diagonal reads
forward or backward.
>>
Am trying to think of a nice non-commutative group of order 16. Here's one:
G = { az + b | a in {1,i,-1,-i}, b in Z[i]/2Z[i] }.
Here b is uniquely = one of 0,1,i,1+i mod 2Z[i].
(Imagine a square torus T tiled by 4 congruent subsquares. Then G is
the group of oriented rigid motions of T that preserve subsquares.)
Then define g o h via (az+b)o(cz+d) = (ac)z + (ad+b)
where ad+b is represented by its remainder mod 2Z[i] = one of 0,1,i,1+i.
Thus 1/(az+b) is given by (1/a)z + -b/a.
Questions:
1. Does this G admit any 4x4 magic squares?
2.If so, is the magic constant uniquely determined?
[A priori we can just assume group multiplication of a row
is left to right, and of a column is top to bottom.]
(I'd guess the most promising magic constant is the identity.
Also, if it's the identity then any cyclic permutation of a given
row or column will multiply to the same thing:
rstu = 1 <=> stur = 1
avoiding the sticky problem of how to define the magic product.)
Hmm, another nice non-commutative group of order 16 is
the isometries of a regular octagon (= D_8). This seems less
interesting than G, but the same questions apply.)
--Dan
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P.S. (I say: diagonals, schmiagonals. Requiring the two diagonals to have
the same magic constant just strikes me as a random rule with little redeeming
recreational math value.)
