# The redundancy of Rubik's cube states [duplicate]

Possible Duplicate:
Rubik’s Cube Not a Group?

I take a Rubik's cube in the solved state, and I secretly assign a unique integer label to each of the cubies. I then, via an arbitrarily long series of random moves, scramble and resolve the cube without looking at the integer labels. Will the integer labels be mapped back to their original positions? Does this answer change for a 4x4x4 cube?

If not, how large is the group of states for this integer labeled cube? Surely it must be quite a bit larger than the Rubik's cube permutation group?

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## marked as duplicate by MJD, hardmath, Ross Millikan, Alexander Gruber♦, Hans LundmarkJan 3 '13 at 8:56

I don't understand the question. When you scramble and re-solve the cube, all the cubies go back to their original position. Since you assigned the integers to the cubies, it seems that the integer must also go back to their original position. Can you clarify? –  Ted Jan 2 '13 at 20:20
@Ted I'm trying to understand if the solved state is unique. –  Yellow Jan 2 '13 at 20:21
How could it not be unique? There's only one solved state, right? I still don't understand. –  Ted Jan 2 '13 at 20:22
In the 3x3x3 case the answer is yes, the solved state is unique. The corners and edges have unique color combinations, leaving only the center square on each face, which is unique in having only the one color. Indeed the face-centered squares are sometimes considered immovable. –  hardmath Jan 2 '13 at 20:24
@Ted There are 54 cubie faces, right? While solving a cube we only consider 6 colors, where each color maps to one of nine possible integer labels. Why is it true that the integers won't be scrambled in the solved state? –  Yellow Jan 2 '13 at 20:24

Based on the comments, it seems like the question is whether the orientation of the center squares is fixed. The Wikipedia page for Rubik's cube says that only half the center orientations ($4^6/2$) are achievable.
Here are instructions for 2 types of center rotations: (1) rotate one center piece clockwise and an adjacent center piece counterclockwise; (2) rotate a single center by 180 degrees. (Corners and edges stay where they are.) Together these generate a subgroup of index 2 inside $(\mathbb{Z}/4\mathbb{Z})^6$ (the subgroup where the sum of all entries is even).