Let (G ; ∗) be
the set of movements of the Rubik's Cube. A movement is defined as follows; one possible
move is a clockwise turn of the top face followed by a counter-clockwise turn
of the right face. If M_1 and M_2 are two moves, then M_1 ∗ M_2 is the move where one does M_1 first, and
then M_2.
PROVE that the set (G ; ∗) is indeed a group under the operation of ∗.
PROVE that the set (G ; ∗) is indeed a group under the operation of ∗.
Proof: I shall now prove this using the axioms of group theory.
(1) It is firstly required
that ∗ be well-defined. It is easy to see this, for if M_1 and M_2
are moves, M_1 ∗ M_2 is a move as well. So G is closed under ∗.
(2) Now, let e be the 'empty' move, i.e. a move that does not change the configuration of the cube in any way. Therefore, M ∗ e means 'first do M, then do nothing'. Therefore, M ∗ e = M. So we have proven that G has an identity, namely, e.
(3) To prove G has an inverse is not taxing in the least. Since M is a move, we can reverse this movement, i.e. do the steps backward to get M', say. Then M ∗ M' simply means 'do M, then reverse the steps of M', i.e. M ∗ M' = e, the empty move. Thus, M' is the inverse of M. Hence it is proved that every element of G has an inverse.
(4) To prove associativity requires a bit more thought and intuition, as well as mechanical sympathy for the structure of the Rubik's cube itself. Upon reflection, and gratuitous analysis, one can see that the cube is made up of 27 smaller cubes of which one cannot be seen. These 'mini' cubes shall be called 'Cubies' for convenience. The space in which these cubies occupy shall be called cubicles. If C is the oriented cubie, write M(C) for the oriented cubicle that C ends up in after the move M is applied, with the faces of M(C) written in the same order as the faces of C. Now, the move M_1 moves C to the cubicle M_1(C); the move M_2 moves M_1(C) to M_2(M_1(C)). Therefore, (M_1 ∗ M_2)(C) = M_2(M_1(C)).
[Recall how moves are defined in the statement above if you are confused]
It is now required that (M_1 ∗ M_2) ∗ M_3 = M_1 ∗ (M_2 ∗ M_3)... (i) be proved, for moves M_1, M_2 and M_3. Upon completion of this, we will have associativity, and the result shall follow... great times.
The proof of (i) above is a direct consequence of proving that [(M_1 ∗ M_2) ∗ M_3](C) = [M_1 ∗ (M_2 ∗ M_3)](C), for any cubie C. Now, [(M_1 ∗ M_2) ∗ M_3](C) = M_3([M_1 ∗ M_2](C)) = M_3(M_2(M_1(C))).
Also, [M_1 ∗ (M_2 ∗ M_3)](C) = (M_2 ∗ M_3)(M_1(C)) = M_3(M_2(M_1(C))).
So (M_1 ∗ M_2) ∗ M_3 = M_1 ∗ (M_2 ∗ M_3). Therefore ∗ is associative.
Thus, finally, by (1), (2), (3) and (4), (G ; ∗) is a group under ∗. ∎
(2) Now, let e be the 'empty' move, i.e. a move that does not change the configuration of the cube in any way. Therefore, M ∗ e means 'first do M, then do nothing'. Therefore, M ∗ e = M. So we have proven that G has an identity, namely, e.
(3) To prove G has an inverse is not taxing in the least. Since M is a move, we can reverse this movement, i.e. do the steps backward to get M', say. Then M ∗ M' simply means 'do M, then reverse the steps of M', i.e. M ∗ M' = e, the empty move. Thus, M' is the inverse of M. Hence it is proved that every element of G has an inverse.
(4) To prove associativity requires a bit more thought and intuition, as well as mechanical sympathy for the structure of the Rubik's cube itself. Upon reflection, and gratuitous analysis, one can see that the cube is made up of 27 smaller cubes of which one cannot be seen. These 'mini' cubes shall be called 'Cubies' for convenience. The space in which these cubies occupy shall be called cubicles. If C is the oriented cubie, write M(C) for the oriented cubicle that C ends up in after the move M is applied, with the faces of M(C) written in the same order as the faces of C. Now, the move M_1 moves C to the cubicle M_1(C); the move M_2 moves M_1(C) to M_2(M_1(C)). Therefore, (M_1 ∗ M_2)(C) = M_2(M_1(C)).
[Recall how moves are defined in the statement above if you are confused]
It is now required that (M_1 ∗ M_2) ∗ M_3 = M_1 ∗ (M_2 ∗ M_3)... (i) be proved, for moves M_1, M_2 and M_3. Upon completion of this, we will have associativity, and the result shall follow... great times.
The proof of (i) above is a direct consequence of proving that [(M_1 ∗ M_2) ∗ M_3](C) = [M_1 ∗ (M_2 ∗ M_3)](C), for any cubie C. Now, [(M_1 ∗ M_2) ∗ M_3](C) = M_3([M_1 ∗ M_2](C)) = M_3(M_2(M_1(C))).
Also, [M_1 ∗ (M_2 ∗ M_3)](C) = (M_2 ∗ M_3)(M_1(C)) = M_3(M_2(M_1(C))).
So (M_1 ∗ M_2) ∗ M_3 = M_1 ∗ (M_2 ∗ M_3). Therefore ∗ is associative.
Thus, finally, by (1), (2), (3) and (4), (G ; ∗) is a group under ∗. ∎
I wish, I wish, I could comprehend in real time.
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