Simple confusion: How can rotation matrix be associative but not abelian? On the Wikipedia page it is said that the rotation matrix is associative but it also states that rotation matrices are not abelian.
The associative property (I think) implies that we have the following condition:
$(R_1R_2)R_3 = R_1(R_2R_3)$
But then this violates the non-abelian condition
Can someone point out the correct associative property of the rotation matrices?
 A: In 2 dimensions, rotation matrices are associative and abelian, which is not hard to visualize. Also, any vector in $\mathbb{R}^{2}$ can be identified in $\mathbb{C}$ as $re^{i\theta}$, and then rotations $R_{\alpha}$, $R_{\beta}$ by angles $\alpha$ and $\beta$, respectively, yields
\begin{equation*}
R_{\alpha}R_{\beta}(re^{i\theta}) = R_{\alpha}(re^{i(\theta + \beta)}) = re^{i(\theta+\beta+\alpha)}
\end{equation*}
Commutativity follows as a result of the commutativity of complex multiplication and the fact that $e^{z_{1} + z_{2}} = e^{z_{1}}e^{z_{2}}$.
However, in 3D things break down. The associativity is true because matrix multiplication in general is associative, regardless of whether we're talking about rotation matrices.
Consider the two rotation operators in $\mathbb{R}^{3}$ given by $R_{\mathbf{z}}$ (rotation by 90 degrees counter clockwise about the z-axis) and $R_{\mathbf{x}}$ (rotation by 90 degrees clockwise about the x-axis). Then consider the image of the vector $\mathbf{e}_{1} := [1,0,0]^{T}$ under a composition of these two operations. It is easy to see that 
\begin{equation*}
R_{\mathbf{x}}R_{\mathbf{z}}\mathbf{e}_{1} = R_{\mathbf{x}}\mathbf{e}_{2} = -\mathbf{e}_{3}.
\end{equation*}
On the other hand,
\begin{equation*}
R_{\mathbf{z}}R_{\mathbf{x}}\mathbf{e}_{1} = R_{\mathbf{z}}\mathbf{e}_{1} = \mathbf{e}_{2}.
\end{equation*}
