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Show that the group of rotations of a cube is isomorphic to $S_4.$

This proof is from Gallian's Abstact Algebra Theorem $7.3$

Proof:Using the Orbit-Stabilizer Theorem we know that the group of rotations of a cube has the same order as $S_4$ (24 elements), we need only prove that the group of rotations is isomorphic to a subgroup of $S_4$. To this end, observe that a cube has four diagonals and the rotation group induces a group of permutations on the four diagonals. But we must be careful not to assume that different rotations correspond to different permutations. To see that this is so, all we need do is show that all $24$ permutations of the diagonals arise from rotations. Labeling the consecutive diagonals $1,2,3,$ and $4$, it is obvious that there is $90^\circ$ rotation that gives the permutation $\alpha = (1234)$; another $90^\circ$ rotation perpendicular to our first one gives the permutations $\beta=(1423)$. See Figure.So the group of permutations induced by the rotations contains the eight-element subgroup $\{\epsilon,\alpha,\alpha^2,\alpha^3,\beta^2,\beta^2\alpha,\beta^2\alpha^2,\beta^2\alpha^3\}$ and $\alpha\beta$, which has order $3$. Clearly, then, the rotations yield all the $24$ permutations.

Figure

My Doubt :I don't understand how $\beta =(1432)$?$\,\,$ I understand how $\alpha=(1234)$ beacuse in this case you rotate the cube shown by the circular turn and $1$ takes the place of $2$, $2$ takes the place of $3$ and $3$ takes the place of $4$. But this is not so in the case of $\beta$. If you rotate as per the figure then $1$ should go to $3$ right?

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  • $\begingroup$ As I see it, on the left side 2 takes the place of 2, 3 of 1, 1 of 4 and 4 of 2. On the right side, the same. Note that you are rotating the cube towards yourself, around an axis which pierces the left and right sides of the cube. $\endgroup$ – B. Pasternak May 22 '16 at 8:39
  • $\begingroup$ OMG!! I pity myself...The arrow always seemed as if it is rotating against me.....thanks alot @B.Pasternak $\endgroup$ – Bijesh K.S May 22 '16 at 8:41
  • $\begingroup$ Don't kid yourself, I saw the arrow like you did the first time I looked at it. There are better ways to depict that arrow. $\endgroup$ – B. Pasternak May 22 '16 at 8:42
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Yes that arrow is bad. You can use $\odot$ and $\otimes$ to denote arrow against the viewer or arrow forwards.

$\odot$ being the tip of a dart facing you (arrow pointing your direction)

$\otimes$ being you seeing the feathers of the dart (vector) as you throw it forward from your p.o.v.

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