"An expert is someone who knows some of the worst mistakes
that can be made in his subject, and how to avoid them.
Heisenberg, Werner
PHYSICS AND BEYOND, 1971
STRUCTURE OF THE RUBIK'S CUBE GROUP
This section shall survey, without proofs, some results on the
grouptheoretical structure of some of the permutation puzzle
groups. We shall follow [GT] (see also [B], chapter 2, [NST],
chapter 19) in our discussion of the pyraminx (tetrahedron), the
3x3 Rubik's cube, and the megaminx (dodecahedron). Other puzzle
groups are analyzed in [GT].
Notation: Let
* Gp (resp., GR, Gm) denote the permutation puzzle group
generated by the basic moves of the pyraminx (resp.,
the Rubik's cube, megaminx),
* Vp (resp., VR, Vm) denote the set of vertex pieces of the
pyraminx (resp., the Rubik's cube, megaminx),
* Ep (resp., ER, Em) denote the set of edge pieces of the
pyraminx (resp., the Rubik's cube, megaminx),
* Fp (resp., FR, Fm) denote the set of facets of the
movable pieces of the pyraminx (resp., the Rubik's cube,
megaminx).
General remarks
Let G,V,E,F (resp.) denote either (Gp,Ep,Vp,Fp),
(GR,ER,VR,FR), or (Gm,Em,Vm,Fm).
Lemma: G acts on the sets V, E, F (individually).
If g is any move in G then, since g acts on the sets
V, E, and F, we may regard g
as an element of the symmetric group S_V of V,
as an element of the symmetric group S_E of E, or
as an element of the symmetric group S_F of F.
These groups S_V, S_E, and S_F are different, so to distinguish
these three ways of regarding g, let us write
g_V for the element of S_V corresponding to g,
g_E for the element of S_E corresponding to g,
g_F for the element of S_F corresponding to g.
Lemma: (a) The function
f_V : G > S_V
g > g_V
is a homomorphism.
(b) The function
f_E : G > S_E
g > g_E
is a homomorphism.
(c) The function
f_F : G > S_F
g > g_F
is a homomorphism.
What is the kernel of f_V? What is its image? To answer this question
(actually, we shall not answer this precise question but one similar to it)
we introduce a subgroup of the symmetric group.
Definition: Let S_X denote the symmetric group of a finite set X. Label
the elements of X as X = { x1, x2, ..., xn } and, using this labeling,
identify S_X with Sn. Let A_X denote the set of all permuations of X which
are even as an element of Sn (in the sense of chapter 2 above). This
set is called the "alternating group" of X.
Lemma: The alternating group is a group.
Aside: The following remarkable result about the alternating group
will not be needed but it is (belive it or not) connected with the
fact that you cannot solve the general polynomial of degree 5 or
higher using radicals.
Theorem: If X has 5 elements or greater then A_X has no nontrivial
proper normal subgroups. In other words, if H < A_X then either
H = {1} or H = A_X.
end aside.
Parity conditions:
Consider the function
f_VE : G > S_VxS_E
g > (g_V,g_E)
It is easy to check that this is a homomorphism.
Theorem: The image f_VE(G) of f_VE is isomorphic to
/ A_VxA_E, for the pyraminx or megaminx,

\ { (x,y) in S_VxS_E  x,y are both even or both odd }, for Rubik's cube.
To see what this means, we look at an example.
Example: Let G=GR. Can you find a move of the Rubik's cube which
flips a single edge subcube over, leaving the rest of the puzzle
pieces unmoved? If so, then the image of f_EV would have to contain
an element (x,y) with x=1 (since moving an edge only does not effect
the vertices) and where y is a 2cycle. But x=1 is even and a 2cycle
is odd. This contradicts the theorem, which says that x,y are either
both even or both odd. Therefore, the answer is no: a single edge
flip is impossible. (Incidently, this is wellknown to cube experts.)
Exercise: Is there a move on the Rubik's cube which flips two corner
subcubes over, without rotation (so it is a 2cycle on the edges connected
to the corner vertices being swapped), leaving the rest of the puzzle
pieces unmoved?
Next, some more notation: let
K = ker(f_VE) < G
denote the kernel of f_VE. This is a normal subgroup of G.
Example: In the case of the Rubik's cube, this is the set of moves
which may reorient (i.e., flip or rotate) a subcube but does
not swap it with some other subcube. For example, the move
((D^2)^R*(U^2)^B)^2,
which twists the ufr corner clockwise and the bld corner counterclockwise,
belongs to K. (Here x^y = y^(1)*x*y, x^2=x*x.)
Theorem: G is a semidirect product of K with f_VE(G):
G = K x f_VE(G).
This is the desired determination of the structure of the permutation
puzzle group. In the case of the 3x3 Rubik's cube, some sketchy details
are given in the next chapter. See also [GT] or [NST] chapter 19.