**Defining the Megaminx Group**

The Megaminx is the shape of a dodecahedron. A dodecahedron is a 12-sided regular platonic solid for which each of the 12 faces is a pentagon. (Platonic solids are the five regular polyhedrons that have "Platonic" groups, or symmetries.) Two faces are neighboring if they share an edge. There are 20 vertices and 30 edges on a dodecahedron. There are a total of 11*12=132 facets on the puzzle. Each face of the solid is parallel to a face on the opposite side. Fix a face of the dodecahedron and consider a plane parallel to that face slicing through the solid and about one-fifth of the way to the opposite face; there are 12 such slices. Two of these slices associated to two neighboring faces will intersect inside the solid, but two such slices associated to two non-neighboring, but non-parallel, faces will not intersect inside the solid; instead, they will intersect outside the solid. This slicing creates a smaller dodecahedron in the center and several other irregular smaller pieces.

For each such slice associated to a given face *f _{i}*, there is a basic move, still denoted

*f _{1}* = (1 3 5 7 9)(2 4 6 8 10)(20 31 42 53 64)(19 30 41 52 63)(18 29 40 51 62)

*f _{2}* = (12 14 16 18 20)(13 15 17 19 21)(1 60 73 84 31)(3 62 75 86 23)(2 61 74 85 32)

*f _{3}* = (23 25 27 29 31)(24 26 28 30 32)(82 95 42 3 16)(83 96 43 4 17)(84 97 34 5 18)

*f _{4}* = (34 36 38 40 42)(35 37 39 41 43)(27 93 106 53 5)(28 94 107 54 6)(29 95 108 45 7)

*f _{5}* = (45 47 49 51 53)(46 48 50 52 54)(38 104 117 64 7)(39 105 118 65 8) (40 106 119 56 9)

*f _{6}* = (56 58 60 62 64)(57 59 61 63 65)(49 115 75 20 9)(50 116 76 21 10) (51 117 67 12 1)

*f _{7}* = (67 69 71 73 75)(68 70 72 74 76)(58 113 126 86 12)(59 114 127 7 13) (60 115 128 78 14)

*f _{8}* = (78 80 82 84 86)(79 81 83 85 87)(71 124 97 23 14)(72 125 98 24 15) (73 126 89 25 16)

*f _{9}* = (89 91 93 95 97)(90 92 94 96 98)(80 122 108 34 25)(81 123 109 35 26) (82 124 100 36 27)

*f _{10}* = (100 102 104 106 108)(101 103 105 107 109)(91 130 119 45 36) (92 131 120 46 37)(93 122 111 47 38)

*f _{11}* = (111 113 115 117 119)(112 114 116 118 120)(102 128 67 56 47) (103 129 68 57 48)(104 130 69 58 49)

*f _{12}* = (122 124 126 128 130)(123 125 127 129 131)(100 89 78 69 111) (101 90 79 70 112)(102 91 80 71 113)

The solid's 12 faces are labeled *f _{1}*,

**Defining a Move**

Let *M* = <*f _{1}*,

Let *V* denote the set of vertices of the dodecahedron, and let D:*H* ->* S _{V}* denote the
homomorphism which associates to each move of the Megaminx the corresponding permutation
of the vertices.

Let *E* denote the set of edges of the dodecahedron, and let F:*H* -> *S _{E}* denote the
homomorphism which associates to each move of the Megaminx the corresponding permutation
of the edges.

For *m* in *M*, let D(*m*) be the permutation of the set of vertices *V* of the solid, and let F(*m*) be
the corresponding permutation of the set of edges *E* of the solid. Let *S _{n} *denote the symmetric
group on

(a) D: *M* -> *S*_{20} is a homomorphism.

(b) F: *M* -> *S*_{30} is a homomorphism.

If * M*, *S*_{20} are groups with *_{1} denoting the group operation for *M* and *_{2} denoting the group
operation for *S*_{20}, then for all *a*,*b* in *M*,* *

D(*a**_{1}*b*) = D(*a*)*_{2}D(*b*).

This same argument can be applied for F: *M* -> *S*_{30}.

In order to mathematically describe the moves of the Megaminx, the corners and the
edges must be oriented as in figures 1-4. Beginning with a solved Megaminx, label the facets
with an invisible "+" (i.e., mark the spatial position of the facet on the cube with a "+"). In the
figures, each facet that receives a "+" is numbered so that the effect of a move *m* on each
particular facet can be observed. These "+" signs are called the "standard reference markings."
Each move *m* of the Megaminx yields a new collection of "+" labels, called the "markings
relative to *m*." A position of the Megaminx is determined by answering the following questions:

(a) How are the edges permuted?

(b) How are the centers permuted?

(c) How are the corners permuted?

(d) Which of the relative edge markings are flipped (relative to the standard reference

markings)?

(e) Which of the relative corner markings are rotated from the standard reference markings and, if so, by how much
(120 or 240 degrees clockwise, relative to the standard reference markings)?

**Corner Orientations**

Let v: *H* -> *C*_{3}^{20} (where *C*_{3}^{20} = *C*_{3}X*C*_{3}X . . . X*C*_{3} (twenty times), and *C*_{3} denotes the cyclic
group of order 3, with addition mod 3 as the operation) be the function which associates each
move *m* in *H* with the corresponding corner orientations.
Let *m* in *H* and say *m* moves corner *i* to
corner *j*. Then v* _{i}* (

(a) a permutation D(*m*) in
*S _{V}* of the vertices

(b) a reorientation of the vertices moved in (a).

Now, using *m,n* in *H*, we must verify that the "relative" orientation v (*mn*) - v (*m*) is the
same as the orientation v (*n*), provided that the effect of *m* on the vertices is taken into account:

v (*n*) = D(*m*)(v (*mn*) - v (*m*)), i.e., v (*mn*) = v (*m*) + D(*m*)^{-1}( v (*n*)).

A sketch of the proof:

The move *mn* orients the *i*^{th} corner subcube by v_{i}(*mn*) and permutes the vertices by D(*mn*),
by definition. On the other hand, *mn* will first act by *m*, then *n*. The move *m* will reorient the *i*^{th}
corner subcube by v_{i}(*m*) and send the *i*^{th} vertex to the D(*m*)(*i*)^{th} vertex. Subtract v (*m*) from v (*mn*)
to study the effect of *n* on this so that the original orientation is back. (This is called the
__modified__ position.) The move *n* firsts orients the *j*^{th} corner subcube of the modified solid by v_{i}(*n*)
and then permutes it to vertex D(*n*)(*j*). The *i*^{th} subcube of the modified solid comes from (via *m*)
the

D(*m*)^{-1}(*i*)^{th} subcube of the original solid. Therefore the *i*^{th} corner subcube of the modified cube is,
by means of *n*, reoriented by vD(*m*)^{-1}(*i*) (*n*). Now add v_{i}(*m*) to get the total effect of *mn* on the *i*^{th}
vertex of the original:

v_{i}(*mn*) = v_{i}(*m*) + vD(*m*)^{-1}(*i*) (*n*) for each 1<*i*<20.

This implies v (*mn*) = v (*m*) + D(*m*)^{-1}( v (*n*)).

**Edge Orientations**

Let w: *H* -> *C*_{2}^{30} (where *C*_{2}^{30} = *C*_{2}X*C*_{2}X . . . X*C*_{2} (thirty times), and *C*_{2} denotes the cyclic
group of order 2, with addition mod 2 as the operation) be the function which associates each
move *m* in
*H* with the corresponding edge orientations.
Let *m* in *H* and say *m* moves
edge *i* to edge *j*.
Then w* _{i}* (

(a) a permutation F(*m*) in
*S _{E}* of the edges

(b) a reorientation of the edges moved in (a).

Now, as with D, we must verify, using * m,n* in
*H*, that the "relative" orientation

w (*mn*) - w (*m*) is the same as the orientation w (*n*), provided that the effect of *m* on the edges is
taken into account:

w (*n*) = F(*m*)(w (*mn*) -w (*m*)), i.e., w (*mn*) = w (*m*) + F(*m*)^{-1}(w (*n*)).

A sketch of the proof:

The move *mn* orients the *i*^{th} edge subcube by w_{i}(*mn*) and permutes the edges by F(*mn*), by
definition. On the other hand, *mn* will first act by *m*, then *n*. The move *m* will reorient the *i*^{th}
edge subcube by w_{i}(*m*) and send the *i*^{th} edge to the F(*m*)(*i*)^{th} edge. Subtract w (*m*) from w (*mn*) to
study the effect of *n* on this so that the original orientation is back. The move *n* firsts orients the
*j*^{th} edge subcube of the modified solid by w_{i}(*n*) and permutes it to edge F(*n*)(*j*). The *i*^{th} subcube of
the modified solid comes from (via *m*) the F(*m*)^{-1}(*i*)^{th} subcube of the original solid. Therefore the
*i*^{th} edge subcube of the modified solid is, by means of *n*, reoriented by wF(*m*)^{-1}(*i*) (*n*). Now add w_{i}(*m*)
to get the total effect of *mn* on the *i*^{th} edge of the original:

w_{i}(*mn*) = w_{i}(*m*) + wF(*m*)^{-1}(*i*) (*n*) for each 1<*i*<30.

This implies w(*mn*) = w (*m*) + F(*m*)^{-1}( w (*n*)).

**The Fundamental Theorem of the Megaminx**

Now, using previously defined functions and homomorphisms, each *m*in
*M* can be
identified with a 4-tuple (D(*m*),F(*m*),v(*m*),w(*m*)).

Given a 4-tuple (r,s,v,w) where:

r in *S*_{20} is a permutation on the corners,

s in *S*_{30} is a permutation on the edges, and

v in *C*_{3}^{20} , w in *C*_{2}^{30} ,

there must be conditions on r,s,v, and w to insure that (r,s,v,w) corresponds to a possible position
of the Megaminx. (In the section on corner and edge orientations, it was already shown that
(r,s,v,w) is at least an impossible position of the Megaminx, since it is simply a combination of
permutations on the edges and corners and reorientations of the edges and vertices.)

Now, it is possible to state

** The Fundamental Theorem of the Megaminx **:

A 4-tuple (r,s,v,w) (r0*S*_{20},
s0*S*_{30},
v0*C*_{3}^{20},
w0*C*_{2}^{30})
corresponds to a possible position of the Megaminx if and only if:

(a) *sgn*(r) = *sgn*(s) = 1 (equal parity of permutations)

(b) v_{1} + v_{2} + . . . + v_{20} /0 (mod 3) (conservation of total twists)

(c) w_{1} + w_{2} + . . . + w_{30} /0 (mod 2) (conservation of total flips)

**Proof:**

__The "only if" part__:

As stated before, each (r,s,v,w)0*S*_{20}**S*_{30}**C*_{3}^{20}**C*_{2}^{30} uniquiely determines a possibly illegal
position of the Megaminx.

(a) Let *m* in *M* be the element which moves the Megaminx from the solved position to
the position associated with this 4-tuple. Then r =D(*m*) and s = F(*m*). The element *m* may be
written as a word in the basic moves *f _{1}*,

*sgn* : *S _{n}* ->{1,-1}
is the homomorphism which associates to a permutation either 1, if it is even,

or -1, if it is odd. Also, *A _{n} = ker*(

homomorphisms, it follows that

*sgn*(r) = *sgn *(D(*m*)) = J*sgn *(D(*X _{i}* )) = J

This proves (a).

(b) To show this, each possible face move, and several "words" from these moves
were performed, (Here, f_{i}* ^{n}* is

X | v(X) |
---|---|

f_{1}
| (0^20) |

f_{2}
| (1,1,0^5,1,0^12) |

f_{3}
| (0^2,1,0^4,1,0,1,0^10) |

f_{4}
| (0^3,1,0^5,1,0,1,0^8) |

f_{5}
| (0^4,1,0^6,1,0,1,0^6) |

f_{6}
| (1,0^12,1,1,0^5) |

f_{7}
| (0^6,1,0^7,2,1,0^3,2) |

f_{8}
| (0^6,2,1,0^10,2,1) |

f_{9}
| (0^8,2,1,0^7,2,1,0) |

f_{10}
| (0^10,2,1,0^4,2,1,0^2) |

f_{11}
| (0^12,2,1,0,2,1,0^3) |

f_{12}
| (0^20) |

f_{1}^{-1}
| (0^20) |

f_{2}^{-1}
| (1,1,0^3,1,0^14) |

f_{1}*f_{2}
| (1,0^3,1,0^2,1,0^12) |

f_{1}*f_{2}^{-1}
| (1,0^3,1,1,0^14) |

f_{1}^{-1}*f_{2 }
| (0,1,1,0^4,1,0^12) |

f_{1}^{-1}*f_{2}^{-1 }
| (0,1,1,0^2,1,0^14) |

f_{2}*f_{1}
| (1,1,0^5,1,0^12) |

f_{2}*f_{1}^{-1}
| (1,1,0^5,1,0^12) |

f_{2}^{-1}*f_{1}
| (1,1,0^3,1,0^14) |

f_{2}^{-1}*f_{1}^{-1}
| (1,1,0^3,1,0^14) |

(f_{1}^{-1}*f_{2}^{-1}*f_{1}*f_{2})^{3} = [f_{1},f_{2}]^{3}
| (2,1,2,0^2,1,0^14) |

m=f_{3}*f^{-1}*f_{4}*f_{2}^{-1}*f_{5}*f_{3}^{-1}*f_{6}*f_{4}^{-1}*f_{2}*f_{5}^{-1} ;
(actual move) f_{6}*f_{1}*m^{6}*f_{1}^{-1}*f_{6}^{-1}
| (0^20) |

f_{1}*f_{6}*f_{1}^{-1}*f_{2}*f_{1}*f_{6}^{-1}*f_{1}^{-1}*f_{2}^{-1}
| (0^5,2,0^8,1,0^5) |

f_{6}*f_{2}*f_{1}*f_{2}^{-1}*f_{1}^{-1}*f_{6}^{-1}*f_{3}^{-1}*f_{1}^{-1}*f_{2}^{-1}*f_{1}*f_{2}*f_{3}
| (0^20) |

(f_{6}^{-1}*f_{2}^{-1}*f_{3}^{-1}*f_{6}*f_{2}*f_{3})^{6}
| (2,2,0^2,2,0^15) |

f_{3}^{-2}*f_{6}^{2}*f_{2}*f_{1}^{-1}*f_{6}*f_{1}*f_{3}^{2}*f_{6}^{2}*f_{1}*f_{6}^{-2}*f_{3}^{-2}*f_{1}^{-1}*f_{6}^{-1}*f_{1}*f_{2}^{-1}*f_{6}^{-2}*f_{3}^{2}*f_{1}^{-1}
| (0^20) |

f_{6}^{-1}*f_{2}^{-1}*f_{1}*f_{3}^{-1}*f_{1}^{-1}*f_{3}*f_{2}*f_{6}*f_{3}*f_{2}*f_{1}^{-1}*f_{6}*f_{1}*f_{6}^{-1}*f_{2}^{-1}*f_{3}^{-1}
| (0^20) |

It should also be shown that the conservation of total twists holds for the composition of
some of these moves. For instance, when composing the moves (f_{6}^{-1}*f_{2}^{-1}*f_{3}^{-1}*f_{6}*f_{2}*f_{3})^{6} and

f_{1}*f_{6}*f_{1}^{-1}*f_{2}*f_{1}*f_{6}^{-1}*f_{1}^{-1}*f_{2}^{-1}, the reorientation of the corners due to the first move is (2,2,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0) while the reorientation due to the second move is

(0,0,0,0,0,2,0,0,0,0,0,0,0,0,1,0,0,0,0,0); the vector sum of these two vectors is: (mod 3)

(2,2,0,0,2,2,0,0,0,0,0,0,0,0,1,0,0,0,0,0). This sum is 9, which is equal to 0 (mod 3).

Additionally,

(i) the conservation of twists condition in (b) is true for (v_{1}, . . . ,v_{20}) if and only if it is
true for any permutation (vD_{(1)}, . . . ,vD_{(20)}),

(ii) as shown above, if (v_{1}, . . . ,v_{20}) and (v'_{1}, . . . ,v'_{20}) each satisfy the conservation of twists condition in (b) then their sum also satisfies it.

As above, say *m* = *X _{1} . . . X_{k}, *where each

v(*X _{1} . . . X_{k-1}X_{k}*) = D(

The term D(*X _{1} . . . X_{k-1}*)

(c) To show conservation of total flips, each possible face move, and several "words" from these moves were performed, and their sums were shown to be 0 (mod 2) as follows:

X | w(X) |
---|---|

f_{1} | (0^30) |

f_{2} | (1,1,0^28) |

f_{3} | (0,1,0,1,0^26) |

f_{4} | (0^3,1,0,1,0^24) |

f_{5} | (0^5,1,0^9,1,0^14) |

f_{6} | (0^8,1,0^2,1,0^18) |

f_{7} | (0^10,1,0^15,1,0^3) |

f_{8} | (0^24,1,1,0^4) |

f_{9} | (0^21,1,0^7,1) |

f_{10} | (0^16,1,0,1,0^11) |

f_{11} | (0^10,1,0^5,1,0^13) |

f_{12} | (0^30) |

f_{1}^{-1} | (0^30) |

f_{2}^{-1} | (1,0^8,1,0^20) |

f_{1}*f_{2}
| (0,1,0^6,1,0^21) |

f_{1}*f_{2}^{-1}
| (0^8,1,1,0^20) |

f_{1}^{-1}*f_{2}
| (0,1,1,0^27) |

f_{1}^{-1}*f_{2}^{-1}
| (0^2,1,0^6,1,0^20) |

f_{2}*f_{1}
| (1,1,0^28) |

f_{2}*f_{1}^{-1}
| (1,1,0^28) |

f_{2}^{-1}*f_{1}
| (1,0^8,1,0^20) |

f_{2}^{-1}*f_{1}^{-1}
| (1,0^8,1,0^20) |

(f_{1}^{-1}*f_{2}^{-1}*f_{1}*f_{2})^{3} = [f_{1},f_{2}]^{3}
| (0^30) |

m=f_{3}*f^{-1}*f_{4}*f_{2}^{-1}*f_{5}*f_{3}^{-1}*f_{6}*f_{4}^{-1}*f_{2}*f_{5}^{-1} ; (actual move) f_{6}*f_{1}*m^{6}*f_{1}^{-1}*f_{6}^{-1}
| (0^30) |

f_{1}*f_{6}*f_{1}^{-1}*f_{2}*f_{1}*f_{6}^{-1}*f_{1}^{-1}*f_{2}^{-1}
| (0^9,1,0,1,0^18) |

f_{6}*f_{2}*f_{1}*f_{2}^{-1}*f_{1}^{-1}*f_{6}^{-1}*f_{3}^{-1}*f_{1}^{-1}*f_{2}^{-1}*f_{1}*f_{2}*f_{3}
| (1,0^8,1,0^20) |

(f_{6}^{-1}*f_{2}^{-1}*f_{3}^{-1}*f_{6}*f_{2}*f_{3})^{6}
| (0^30) |

f_{3}^{-2}*f_{6}^{2}*f_{2}*f_{1}^{-1}*f_{6}*f_{1}*f_{3}^{2}*f_{6}^{2}*f_{1}*f_{6}^{-2}*f_{3}^{-2}*f_{1}^{-1}*f_{6}^{-1}*f_{1}*f_{2}^{-1}*f_{6}^{-2}*f_{3}^{2}*f_{1}^{-1}
| (1,0^7,1,0^21) |

f_{6}^{-1}*f_{2}^{-1}*f_{1}*f_{3}^{-1}*f_{1}^{-1}*f_{3}*f_{2}*f_{6}*f_{3}*f_{2}*f_{1}^{-1}*f_{6}*f_{1}*f_{6}^{-1}*f_{2}^{-1}*f_{3}^{-1}
| (1,0,1,0^27) |

As before, it should be shown that the conservation of total flips holds for the
composition of some of these moves. Composing again the moves (f_{6}^{-1}*f_{2}^{-1}*f_{3}^{-1}*f_{6}*f_{2}*f_{3})^{6} and

f_{1}*f_{6}*f_{1}^{-1}*f_{2}*f_{1}*f_{6}^{-1}*f_{1}^{-1}*f_{2}^{-1}, the reorientation of the edges due to the first move is

(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,)

while the reorientation due to the second move is

(0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0);

the vector sum of these two vectors is: (mod 2)

(0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0).

This sum is 2, which is equal to 0 (mod 2).

Additionally,

(i) the conservation of flips condition in (c) is true for (w_{1}, . . . ,w_{30}) if and only if it is true for any permutation (wF_{(1)}, . . . ,wF_{(30)}),

(ii) as shown above, if (w_{1}, . . . ,w_{30}) and (w'_{1}, . . . ,w'_{30}) each satisfy the conservation of
flips condition in (c) then their sum also satisfies it.

As before, in the previous listing of some w(*X*)'s, all words of length *k* = 1 were checked for
conservation of flips. Assume *k* > 1. Using the formula giving the orientation of the product of
two moves in terms of the two orientations of the moves, it can be shown that

w(*X _{1} . . . X_{k-1}X_{k}*) = F(

The terms F(*X _{1} . . . X_{k-1}*)

**Proof:**

** **__The "if" part:__

__ __Assuming (a), (b), and (c) are true, it needs to be shown that there is a corresponding legal
position of the Megaminx. This will be a constructive proof, built by using several special cases.

CASE 1

Suppose that, given the 4-tuple (*r,s,v,w*), both *r* and *s* are the identity and no edges have
been flipped; i.e. (w_{1}, . . . ,w_{30}) = (0, . . . ,0). The claim is that there exists an *m*
in *M *such that

D(*m*) = 1 , F(*m*) = 1 , w(*m*) = (0, . . . ,0) , and v(*m*) = (v_{1}, . . . ,v_{20}).

It can be shown that, for any *i,j*0 {1, . . . ,20}, *i* = *j*, there is an
* m* in *M *such that D(*m*) = 1 ,
F(*m*) = 1 , w(*m*) = (0, . . . ,0) , and v(*m*) = (v_{1}, . . . ,v_{20}) , where v_{i}= 1, v_{j}= 2, and v_{k}= 0, * k *= *i,j*.

This calls for a move which permutes neither the corners nor the edges, flips no edges, and twists
one corner clockwise (a value for the orientation of 1), and one corner clockwise twice (the same as
counterclockwise once, or a value for the orientation of either (-1) or 2). The conservation of total
twists is seen to be preserved in that v_{1}+ . . . +v_{20 } = 0 (mod 3). Such a move *m* is known to exist.
For instance, performing the move

*m* = (f_{6}^{-1}*f_{2}^{-1}*f_{3}^{-1}*f_{6}*f_{2}*f_{3})^{6}*f_{6}*(f_{6}^{-1}*f_{1}^{-1}*f_{4}^{-1}*f_{6}*f_{1}*f_{4})^{12}*f_{6}^{-1}

causes a clockwise twist of the (f_{1}/f_{4}/f_{5 }) corner and a counterclockwise twist of the (f_{1}/f_{2}/f_{3}) corner.

This proves the "if" part of the theorem in the case that *r* and *s* are both the identity and that

(w_{1}, . . . ,w_{30}) = (0, . . . ,0).

CASE 2

Suppose that, given the 4-tuple (*r,s,v,w*), both *r* and *s* are the identity and no corners have
been twisted; i.e. (v_{1}, . . . ,v_{20}) = (0, . . . ,0). The claim is that there exists an
*m* in *M *such that

D(*m*) = 1 , F(*m*) = 1 , v(*m*) = (0, . . . ,0) , and w(*m*) = (w_{1}, . . . ,w_{30}).

It can be shown that, for any *i,j*0 {1, . . . ,30}, *i* = *j*, there is an
* m* in *M *such that D(*m*) = 1 ,
F(*m*) = 1 , v(*m*) = (0, . . . ,0) , and w(*m*) = (w_{1}, . . . ,w_{30}) , where w_{i}= 1, w_{j}= 1, and w_{k}= 0, * k *= *i,j*.

This calls for a move which permutes neither the corners nor the edges, twists no corners, and flips
only two edges (a value of 1 for each of the orientations for the edges). The conservation of total
flips is seen to be preserved in that w_{1}+ . . . +w_{30 } = 0 (mod 2). Such a move *m* is known to exist.

The move *m* = f_{6}^{-1}*f_{2}^{-1}*f_{1}*f_{3}^{-1}*f_{1}^{-1}*f_{3}*f_{2}*f_{6}*f_{3}*f_{2}*f_{1}^{-1}*f_{6}*f_{1}*f_{6}^{-1}*f_{2}^{-1}*f_{3}^{-1} flips only the (f_{1}/f_{2}) and (f_{1}/f_{3})
edges. This proves the "if" part of the theorem in the case that *r* and *s* are both the identity and that

(v_{1}, . . . ,v_{20}) = (0, . . . ,0).

CASE 3

Finally, given the 4-tuple (*r,s,v,w*), assume that both (w_{1}, . . . ,w_{30}) = (0, . . . ,0) and

(v_{1}, . . . ,v_{20}) = (0, . . . ,0); i.e. no edges have been flipped and no corners have been twisted. The
claims are that

(a) given any three edge subcubes, there is a move *m* in *M* which is a 3-cycle on these
edges

(b) given any three corners, there is a move *m* in *M* which is a 3-cycle on these corners

and (c) given any three edges and any three corners, there is a move *m* in *M* which is a 3-cycle

on these edges and a 3-cycle on these corners.

As stated in the beginning of this paper, let *H* be the subgroup of *S _{n}* generated by all the 5-cycles in

(a) This calls for a move which is a 3-cycle on any three given edges and where *sgn* (*s*) =
1. Such a move *m* is known to exist. The move *m* = f_{6}*f_{2}*f_{1}*f_{2}^{-1}*f_{1}^{-1}*f_{6}^{-1}*f_{3}^{-1}*f_{1}^{-1}*f_{2}^{-1}*f_{1}*f_{2}*f_{3} cause
a

3-cycle on the (f_{1}/f_{2}), (f_{2}/f_{3}) and (f_{2}/f_{6}) edges.

(b) This calls for a move which is a 3-cycle on any three given corners and where

*sgn* (*r*) = 1. Such a move *m* is known to exist. The move

*m* = f_{3}*f_{8}^{2}*f_{7}^{2}*f_{1}*f_{6}*f_{1}^{-1}*f_{2}*f_{1}*f_{6}^{-1}*f_{1}^{-1}*f_{2}^{-1}*(f_{6}*f_{2}*f_{1}*f_{2}^{-1}*f_{1}^{-1}*f_{6}^{-1}*f_{3}^{-1}*f_{1}^{-1}*f_{2}^{-1}*f_{1}*f_{2}*f_{3})^{2}*f_{7}^{2}*f_{8}^{2}*f_{3}

is a 3-cycle on the (f_{1}*f_{2}*f_{6}), (f_{1}*f_{7}*f_{6}), and (f_{2}*f_{6}*f_{7}) corners.

(c) This calls for a move which is a 3-cycle on any three given edges and a 3-cycle on three
corners; *sgn* (*s*) = *sgn* (*r*) = 1. Such a move *m* is known to exist. The move

*m* = f_{1}*f_{6}*f_{1}^{-1}*f_{2}*f_{1}*f_{6}^{-1}*f_{1}^{-1}*f_{2}^{-1} is a 3-cycle on the (f_{1}/f_{2}), (f_{6}/f_{7}) and (f_{2}/f_{6}) edges and a 3-cycle on the
(f_{1}*f_{2}*f_{6}), (f_{1}*f_{7}*f_{6}), and (f_{2}*f_{6}*f_{7}) corners.

This constructively proves the "if" part of the proof.

**The Size of M**

Given *H*, the group of all illegal moves not affecting the centers of the Megaminx group,

*H* = *C*_{3}^{20}X
*S*_{20}X
*C*_{2}^{30}X
*S*_{30},
*M* is a subgroup of *H*. Because of the conditions set forth in the Fundamental
Theorem (conservation of total twists and flips), an extra condition is imposed on the vectors v and
w as follows:

Since v(*m*) = (0, . . . ,0) there are only seven degrees of freedom, and v_{1}+ . . . +v_{19}
= -v_{20}.

It also follows that since w(*m*) = (0, . . . ,0),
w_{1}+ . . . +w_{29}
= -w_{30}.
Therefore, * M* is actually a
subgroup of a larger group *M'*
= *C*_{3}^{19}X
*S*_{20}X
*C*_{2}^{29}X
*S*_{30}.
Also, by the Fundamental Theorem of the
Megaminx, D(*M*) = *A _{20}*
and F(

In particular,

**M** =
(.5)^{2}
**S*_{20} ***S*_{30}
***C*_{2}^{29}
***C*_{3}^{19} *
= (.5)^{2}20!30!
2^{29}2^{19}.

This number is approximately 1.006 X 10^{68}.

Coreyanne Rickwalt

U.S. Naval Academy

Created 11-21-97.

Last updated 6-7-2004 by wdj.