the group of symmetries of the nodes of a cube Can you explain me please how can  I find rotation for a cube. I will attach an image:

Thanks :) 
 A: started as a comment, but way too long, so posted here as a suggestion to look into
Maybe you have a particular reason to look at permutations of the nodes, but if you want to investigate the rotational symmetry of the cube, it is probably easier to use that its rotational symmetry group is isomorphic to $S_4$. To do this, you won't look at permutations of the vertices, but imagine the 4 diagonals inscribed into the cube, going, eg, from 1 to 7. Each rotational symmetry rotates the 4 diagonals among themselves, and so corresponds to a permutation in $S_4$. All rotations are generated by three types of rotations around axes: (i) through the center of opposing faces (by $\frac{k \pi}{2}$, $k = 0, 1, 2, 3$); (ii) through one of the diagonals (by $\frac{2 \pi}{3}$ and $\frac{4 \pi}{3}$); and (iii) through a line connecting the middle of two opposing edges (eg, middle 12 to middle 67) (by $\pi$). As indicated by a comment by Barto above, the rotation you are particularly interested in is of type (ii), and you can rotate by either $\frac{2 \pi}{3}$ or $\frac{4 \pi}{3}$. This still twists your mind (or at least it does mine, whenever I go through this again), but it is a more parsimonious description of the cube's symmetry.
