Prove that the group of moves of the Rubik’s cube is not abelian.

Question

I'm currently working in the following excercise:

Remember that $G$ is the group of moves of the Rubik’s cube. Prove that this group is not abelian.

I'm starting from picking two moves $M_1$ and $M_2$ and I'm looking at their commutator $[M1, M2]$, which is defined to be $M_1M_2M_1^{−1}M_2^{-1}$, but I'm not sure this is the way to proceed and which steps to go forward with the proof.

Thanks in advance for any hint or help and for taking the time to read my question.

Answer 1

More precisely, $G$ is the set of positions reachable from a solved cube, since the single-turn moves are obviously not closed under addition.

As stated in comments, the proof that $G$ is non-abelian proceeds by counterexample: $[F,R]=FRF'R'\ne e$, using Singmaster notation.

Answer 2

Consider a directed graph whose vertex set $V$ is the set of all configurations of Rubik's cube, and which has an arc (directed edge) from $x$ to $y$ with edge label $A$ if it is possible to go from $x$ to $y$ by applying transformation $A$.

It needs to be shown that if $A$ and $B$ are transformations, then $AB$ is in general not equal to $BA$.

It suffices to show that in the directed graph constructed above, if the current configuration is $z$, then taking the arc $A$ followed by $B$ will give a different destination vertex (i.e. different destination configuration) than if arc $B$ was taken before arc $A$. But this is clear if you visualize the Rubik's cube (or obtain a cube and try from any given starting configuration). For example, take $A$ and $B$ to be 90 degree clockwise rotations of the top face and the right face, respectively.