Here shows a cube with base touching face edges' vertices as: $\{3,7\},\{7,8\}, \{8,4\}, \{4,3\}.$
The opposite face has edges' vertices $\{1,2\},\{2,6\}, \{6,5\}, \{5,1\}.$
It states the below :
$$x = \text{rotation about } \,x = (1, 4, 8, 5) (2, 3, 7, 6)$$ $$y =\text{rotation about}\, y = (1, 5, 6, 2) (4, 8, 7, 3)$$ $$z = \text{rotation about}\, z = (1, 2, 3, 4) (5, 6, 7, 8)$$
Shouldn't be: $$x = \text{rotation about }\, x = (1, 4) (8, 5) (2, 3) (7, 6)$$ $$y =\text{rotation about}\, y = (1, 5) (6, 2) (4, 8) (7, 3)$$ $$z = \text{rotation about} \,z = (1, 2)(3, 4) (5, 6) (7, 8)$$
No.
Consider the square face $(1485)$. It requires a four-cycle for a rotation about the $x$-axis, since, otherwise, two disjoint two-cycles would have order two.
Such a rotation of the square is a rotation of the cube, provided you take into account the opposite square.
The other axes are similar.