This is from visual group theory by Nathan Carter.
It mentions that moves can be combined in any sequence, where move is rotation by 90 degrees of face of rubik cube.
How is that $a(bc) = (ab)c ?$
Why in Rubik cube moves can be combined in any sequence
group-theory
Related Solutions
(This answer only discusses under the context of the MIT paper OP referred to.)
I have done a research on "why the heck did they put commutator and conjugation in Rubik's Cube notes" (written in a more polite language of course) and read tons of materials related to Rubik's Cube.
Basically, there is no theoretical significance now. It is useful for practical reason and stays there for historic and educational reasons.
For people who has attempted solving the Cube, I'm sure it happens at some point of time that you say to yourself "let's put this cubie away using move $A$ so the next move $B$ (which usually does something helpful) does not affect it, and we can bring it back later with a reverse move $A^{-1}$". This is conjugation $ABA^{-1}$. Commutator has similar practical meaning.
They are useful when people want to figure out by themselves how to solve the Cube because they are simple and the effect is obvious to eyes, so they are used extensively in last century when people do not have the computational power to look at the mathematical essence of the Cube. I believe Singmaster's Notes on Rubik's Magic Cube and possibly one issue of Hofstadter's Metamagical Themas that introduces Rubik's Cube have significant influence on this.
Right now the academia is not really concerned about how to solve the Cube anymore since you can easily find some algorithms online. Rubik's Cube is mostly used for demonstrating or introducing concepts in group theory, and given that historically people often see commutator and conjugation in books about Rubik's Cube, it is natural to pick them as two of the concepts that are demonstrated by Rubik's Cube.
Starting from a completed cube, each time you perform your set of moves, you either get a state of the cube that you have seen before, or one that you haven't had before. As there are only finitely many states (even though this is a very big number), you must at some point get to a state you've had before. (Otherwise we'd always be finding a new state so there'd be infinitely many.)
Let's number the steps: 0 is the complete state we start from, applying the moves once gets you to 1, then again to 2, etc.
Then as we saw before, at some step $n$, we get a state we've seen before, say state $m$ (where $m < n$). By applying the reverse of the set of moves from here, we get state $m-1$ which is also state $n-1$. Since this was the first repeated state, we have to have $m = 0$, else $n-1$ would be an earlier repetition.
(By reverse, for instance if we had Down Left we'd do Left' Down', so taking moves back to front and flipping them from clockwise to counter clockwise.)
That means that after $n$ of the sets of moves you already had, we get the original state.
Group theory can help you formalise this, which can help to get the specific numbers (6, 12) for this and potentially other sets of moves.
Best Answer
Association of Rubik moves is so trivial it can be hard to spot.
Consider the move TLF (Top face a quarter turn, then Left face a quarter turn, then Front face a quarter turn). If we first do T, then do LF, we get the same result as we do when we first do TL, then do F. In other words, T(LF) = (TL)F. The Rubik's cube is associative.