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By Exercise 25 every cyclic subset can be factored into cyclic subsets of prime cardinalities. By Exercise 26 if the original cyclic 28 ALGEBRA AND TILING subset is not a subgroup t h e n neither are the new ones; and if none of the new ones is a subgroup then neither is the original. Thus we may suppose that in Hajos's version all the cyclic subsets are of prime orders. But now the n u m b e r of the factors need n o longer correspond to the dimension of the original cube tiling. As we mentioned earlier, Hajos settled Minkowski's conjecture in 1941 using this approach.

As we mentioned earlier, Hajos settled Minkowski's conjecture in 1941 using this approach. 7. Related work In 1930 Keller [6] suggested that the "lattice" condition in Minkowski's problem is irrelevant, conjecturing that in a tiling of η-space by parallel unit cubes there must be twins. Perron in 1940 [16] verified this conjecture for η < 6. If this conjecture were true it would imply Hajos's theorem, and show that Minkowski's t h e o r e m was really geometric, not algebraic. However, in 1992 Lagarias and Shor [9], using an approach of Corrädi and Szabo [2], showed that Keller's conjecture is false in all dimensions greater than or equal to 10.

These tubes, in turn, form slabs, a n d the slabs fill u p 3-space. 22 ALGEBRA AND TILING Exercise 23. Consider a tiling of 3-space by a lattice of unit cubes. Show that the lattice has a basis of the form (1,0,0), ( ΐ ι , Ι , Ο ) , ( x , x , l ) , 2 3 or a basis obtained from this by a permutation of t h e coordinates. This was Minkowski's conjecture, m a d e in algebraic form in 1896 and in geometric form in 1907. Conjecture. In a lattice tiling of η-space by unit cubes there must be a pair of cubes that share a complete (n — l)-dimensional face.

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