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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.

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    $\begingroup$ Get yourself a (solved) rubik's cube and try the moves $RUR'U'$. If the cube isn't solved afterwards, the commutator isn't the identity. It doesn't have to be a real one, it could be a virtual one as well. $\endgroup$ – jgon May 26 at 2:30
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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.

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    $\begingroup$ Hmm but how does $G$ being the group of positions make sense? So $G$ is a set consisting of $100$ trillion positions (or whatever the number is)? How do you compose two positions? $\endgroup$ – Ovi May 26 at 2:44
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    $\begingroup$ @Ovi Positions are represented by the sequence of moves needed to get from the solved state to that position. Positrons are composed by performing their corresponding moves, one sequence after the other. $\endgroup$ – Parcly Taxel May 26 at 2:45
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    $\begingroup$ @Ovi Also, the number of cube positions is 43 quintillion. $\endgroup$ – Parcly Taxel May 26 at 2:49
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    $\begingroup$ Ah thank you very much for explaining. Could I also ask you about clarification on the second part of your first sentence? In my mind, if you fix the Rubik's cube and only allow yourself to twist it's layers, there are 9 twitsts you could do. Is this related to your first sentence? $\endgroup$ – Ovi May 26 at 4:21
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    $\begingroup$ @Ovi Yes, in that there are far fewer results of turn compositions than are positions. $\endgroup$ – Parcly Taxel May 26 at 6:02
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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.

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