Why in Rubik cube moves can be combined in any sequence? 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 $a(bc) = (ab)c ?$
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.
As Arthur stated, its trivial and therefore strangely hard to spot; I came here with the same question. People seem to be misled when trying to think in terms of ordering compositions of transformations. Its not so simple as (F ∗ U) ∗ R where ∗ means "then perform"; because they then think logically to apply PEMDAS to F ∗ (U ∗ R) where parentheses also suggests "do this first, then perform...". It fails because the cube is obviously non-abelian, you can't "F then do UR". This points out the failure of good notation (and therefore explanation). (I think) its made most clear by Janet Chen in her writeup Group Theory and the Rubik’s Cube


  If C is an oriented cubie, we will write M(C) for the oriented cubicle
  that C ends up in after we apply the move M, with the faces of M(C)
  written in the same order as the faces of C. That is, the first face
  of C should end up in the first face of M(C), and so on. For example,
  the move R puts the ur cubie in the br cubicle, with the u face of the
  cubie lying in the b face of the cubicle and the r face of the cubie
  lying in the r face of the cubicle. Thus, we write R(ur) = br. First,
  let’s investigate what a sequence of two moves does to the cubie. If
  M1 and M2 are two moves, then M1 ∗ M2 is the move where we first do M1
  and then do M2. The move M1 moves C to the cubicle M1(C); the move M2
  then moves it to M2(M1(C)). Therefore, (M1 ∗ M2)(C) = M2(M1(C)). To
  show that ∗ is associative, we need to show that (M1 ∗ M2) ∗ M3 = M1 ∗
  (M2 ∗ M3) for any moves M1, M2, and M3. This is the same as showing
  that (M1 ∗ M2) ∗ M3 and M1 ∗ (M2 ∗ M3) do the same thing to every
  cubie. That is, we want to show that (M1 ∗ M2) ∗ M3 = M1 ∗ (M2 ∗
  M3) for any cubie C. We know from our above calculation that (M1
  ∗ M2) ∗ M3 = M3(M1 ∗ M2) = M3(M2(M1(C))). On the other hand,
  M1 ∗ (M2 ∗ M3) = (M2 ∗ M3)(M1(C)) = M3(M2(M1(C))). So, (M1 ∗ M2)
  ∗ M3 = M1 ∗ (M2 ∗ M3). Thus, ∗ is associative.


Hope that helps.