TetraMag Puzzle
The TetraMag puzzle[1] is a collection of 216 powerful steel-magnet spheres, which can be arranged into many configurations. The object of the puzzle is to return to an initial position after disassembling the puzzle. Additional constraints can be imposed, such as while disassembling to not cut the puzzle into more than one pieces, but these vary depending on the objective. The most common configuration is the cube configuration, which is shown on the web page link, above.
Analysis
The puzzle's primary cyclic groups are the groups Z/2mZ=Z2m and Z/3nZ=Z3n, for m=n=3, therefore the puzzle group is a finitely generated abelian group G of order 216=63=23*33, and hence every subgroup and factor group of G is again finitely generated and abelian.
As the direct product of the primary cyclic subgroups Z23 and Z33 with GCD(2,3)=1, G is cyclic.
The units of the ring Z216 are the numbers coprime to 216. They are the set:
(Z/216Z)*=Z216*={1,5,7,11,13,17,19,23,25,29,31,35,37,41,43,47,49,53,55,59,61,65,67,71, 73,77,79,83,85,89,91,95,97,101,103,107,109,113,115,119,121,125,127,131,133,137,139,143, 145,149, 151,155,157,161,163,167,169,173,175,179,181,185,187,191,193,197,199,203,205,209,211,215},
which has cardinality φ(216)=72, and forms a group under multiplication modulo 216.
The order of G is not of the form 2*pk for an odd prime p and k ≥ 1, so not every generator of Z216* is a primitive root modulo 216. In fact, λ(216)=LCM(18,6,9,2,3)=φ(216)/4=18, so Z216* is not cyclic.
The Chinese Remainder Theorem states that Z216* ~ Z23* x Z33* ~ C2 x C2 x C18, whose order is 72=φ(216).
Since G is finitely generated, it satisfies the fundamental theorem of abelian groups, as G ~ Z23 x Z33.
G's cyclic subgroups have order d, with d|216:
> with(numtheory):
> N:=216;
> for m from 0 to 3 do
> for n from 0 to 3 do
> print(`subgroup: Z`,2^m,`+Z`,3^n,`ord`,2^m*3^n,,`index`,N/(2^m*3^n));
> od;
> od;
The above code produces the following table of subgroups:
| subgroup H | order | [G:H] |
| Z1 x Z1 | 1 | 216 |
| Z1 x Z3 | 3 | 72 |
| Z1 x Z9 | 9 | 24 |
| Z1 x Z27 | 27 | 8 |
| Z2 x Z1 | 2 | 108 |
| Z2 x Z3 | 6 | 36 |
| Z2 x Z9 | 18 | 12 |
| Z2 x Z27 | 54 | 4 |
| Z4 x Z1 | 4 | 54 |
| Z4 x Z3 | 12 | 18 |
| Z4 x Z9 | 36 | 6 |
| Z4 x Z27 | 108 | 2 |
| Z8 x Z1 | 8 | 27 |
| Z8 x Z3 | 24 | 9 |
| Z8 x Z9 | 72 | 3 |
| Z8 x Z27 | 216 | 1 |
Solving the Puzzle
Since |G|=216=23*33, Burnside's Theorem guarantees that G is is solvable, so we can find specific subgroup decomposition sequences, which can manually solve the puzzle.
On the video below, one such sequence is shown, which completely decomposes the puzzle and then re-composes it: