A question about why $S^2$ is not a group under given operation. I have a little perplexity reading this document.
Consider the sphere $S^2$ and its north pole $v = (0,0,1)$. We can identify any point on the sphere by angles $(\alpha,\beta)$, literally writing $$ S^2 \ni x = R_z(\alpha)R_y(\beta)v$$
Now we define the operation:
$$x_1 \cdot x_2:=R_z(\alpha_1)R_y(\beta_1)R_z(\alpha_2)R_y(\beta_2)$$
Why this operation is not closed in $S^2$? In other words, why $S^2$ is not a group under this operation?
 A: To quote pages 4–5 of the linked article:

For a negative example, consider the sphere $S^{2}$ and its north pole $v = (0,0, 1)$. We can identify any point on the sphere by angles $(\alpha, \beta)$, which represent a rotation of the north pole $R_{y}(\beta)$ (around $y$) followed by $R_{z}(\alpha)$ (around $z$); we write $x = R_{z}(\alpha) R_{y}(\beta)v$. Now define the operation $\cdot$ such that $x_{1} \cdot x_{2} = R_{z}(\alpha_{1}) R_{y}(\beta_{1}) R_{z}(\alpha_{2}) R_{y}(\beta_{2})$. Any rotation in $SO(3)$ can be represented  as $R_{z}(\alpha_{1}) R_{y}(\beta_{1}) R_{z}(\alpha_{2}) R_{y}(\beta_{2})$, and not only the ones of the form $R_{z}(\alpha_{1}) R_{y}(\beta_{1})$; therefore the operation $\cdot$ as defined is not closed in $S^{2}$, and $(S^{2}, \cdot)$ is not a group.

The explanation is stated explicitly: Not every composition of a rotation about $y$ with a rotation about $z$ is a rotation about $y$ with a rotation about $z$.
To see why this is, we can represent rotations about the axes as $3 \times 3$ matrices and multiply. A short calculation shows that $R_{z}(\alpha_{1}) R_{y}(\beta_{1})$ has $(3,2)$-entry (third row, second column, i.e., the $z$ component of the image of $(0, 1, 0)$ equal to $0$.
This fact is geometrically apparent in retrospect as well: The vector $(0, 1, 0)$ is invariant under rotation about the $y$-axis, and stays in the plane $z = 0$ under rotation about the $z$-axis. Any rotation that sends to $(0, 1, 0)$ to a vector not in the plane $z = 0$ is therefore not an element of $S^{2}$ in this sense. Explicitly,
$$
R_{z}(0) R_{y}(\pi/4) R_{z}(-\pi/2) R_{y}(0) = R_{y}(\pi/4) R_{z}(-\pi/2)
$$
is a composition of two "rotations from $S^{2}$," but the vector $(0, 1, 0)$ is sent to $\frac{1}{2}\sqrt{2}(1, 0, -1)$, which does not lie in the plane $z = 0$, and is therefore not the image of $(0, 1, 0)$ under an element of $S^{2}$.
