A Rubik's Cube has six sides, each of which can be turned clockwise by 90, 180, or 270 degrees. Consider a finite sequence of such turns. (An example of such a sequence is "Right 90, Back 180, Left 270, Front 90, Right 270," where e.g. Right 90 means "turn the Right side 90°.") (a) Start with a solved Rubik's Cube, and apply the same sequence of turns repeatedly. Show that, after some number of applications of this sequence, the Rubik's Cube will again be solved. (Hint: Define a suitable group.)(b) Give an explicit number N such that, for any finite sequence of turns, the procedure of the previous part yields a solved Rubik's Cube in < N applications of the sequence. (This N does not need to be optimal or even remotely close to optimal. You should not need to use any hard facts about the Rubik's Cube. For example, it is enough to know that a Rubik's cube has 9 x 6 = 54 stickers.)

Let L, R, F, B, U, D denote the elementary moves of turning clockwise the left, right, front, back, up and down faces respectively of the Rubik's cube. Also, let I denote the move of doing nothing to the Rubik's cube(Yeah, this sounds strange).

Define a word in L,R,F,B,U,D to be any finite sequence of the six letters. Examples are LRFFDUDUU , LLL , FFLFFB , etc.

Let Moves = { Words in L,R,F,B,U,D } U { I }.

Define a binary operation . on Moves as follows :

Given A,B in Moves, define A.B = AB (the concatenation of A and B) with the rules that :

● LLLL = RRRR = FFFF = BBBB = UUUU = DDDD = I (intuitively, observe that repeating any elementary move(such as L or D or F etc) 4 times sums up to doing nothing to the Rubik's cube, which is the identity move I.

● If A is any element in Moves, then A.I = I.A = A

● Any other relation which is supposed to hold between L,R,F,B,U,D considering an actual Rubik's cube.

Observe that ( Moves, . ) is a group with identity I (doing nothing) and inverse of any element E1E2...En to be En3En-13...E13 where E1,E2,...En are in {L,R,F,B,U,D} . Roughly speaking, (Moves, .) is the group of all moves in a Rubik's cube under the operation of concatenation.

It can be proved combinatorially that this group has 12!2108!37 elements. In fact, observe that if a Rubik's cube is manipulated, the six center cubicles always stay fixed. The rest of the 48 cubicles are permuted. Hence, this group is a subgroup of S48 which is finite. Thus, this group is finite.

Now, any sequence of turns(as described in the first part of the problem) is just an element in the group (Moves,.). Since the group is finite, hence the order of this element is also finite. Consequently, repeating that sequence of turns again and again yields the identity move I(doing nothing to the cube). Consequently, the cube again comes back to the original configuration i.e. it again gets solved.

(b)

Now, any element A in Moves is such that o(A) | |Moves| (by Lagrange's Theorem).

Hence, order of any element in (Moves,.) divides 48!. Thus, for any element A in (Moves,.), A 48! = I.

Hence, let N = 48!

Step-by-step explanation: