# 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 :)

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Could you explain a little more what you're looking for? Is it the number of rotation-symmetries of a cube, or would you like to count all symmetries? – barto Jan 4 '13 at 22:03
@barto I'm looking to find for example the rotation around axis $3-5$. Why the rotations for axis $3-5$ are $(3)(5)(1,8,6) \ldots$ . thanks :) – Iuli Jan 4 '13 at 22:10
Ok. Imagine you rotate over $120$ degrees, so that number $8$ comes on the location where $1$ was before rotating: say $8'=1$. Also, $1'=6$ and $6'=8$. Now look where $4$ comes: $4'=2$, $2'=7$ and $7'=4$. This is why there's written $(3)(5)(1,8,6)(2,4,7)$: Between every pair of brackets in this notation $()$, the numbers are cycled around in the same direction when the cube is rotated: $3\to3$, $5\to5$, $1\to8\to6\to1$ and $2\to4\to7\to2$. I think that's the logic behind this notation (which I have not seen before...) – barto Jan 4 '13 at 22:19
@barto How you make rotations? using only the angle of $120$ degrees? – Iuli Jan 4 '13 at 22:31
@Iuli, if you look at the vertex 3, you see that there are three edges meeting there: 23, 43 and 73. The rotation keeps the vertex 3 fixed, so it has to permute those three edges. That edge permutation actually has to be a 3-cycle (a 2-cycle would be a reflection). So the rotation has order three. Hence it must be a rotation by 120 degrees, and it must permute the other end points of those three edges, 2, 4 and 7, cyclically. – Jyrki Lahtonen Jan 4 '13 at 22:38

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.

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