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The following question is taken from M.A. Armstrong's Groups & Symmetry #17.3. It reads as follows:

Having identified $S_4$ with the rotational symmetry group of the principal diagonals of the cube, consider the action of $A_4$ on the set of vertices of the cube. Calculate the orbits and stabilizers of each orbit.

Currently, I am just trying to determine a quick way of calculating the action of an element of $A_4$ on the vertices. For example, consider the action of $(12)(34) \in A_4$ on the vertices numbered 1 through 8 (vertices 1 through 4 on the top and 5 through 8 on the bottom with 1 above 5, 2 above 6, 3 above 7, and 4 above 8). I first apply $(12)(34)$ to the cube, i.e. swap principal diagonals 3 with 4 followed by 1 with 2, and then record the effect this has on the vertices. In this case, we obtain the following permutation of the vertices: $(18)(27)(36)(45)$.

This is tedious. I feel like there may be a shortcut (say, exploiting the principal diagonals) to this. Any insights are appreciated.

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Instead of building the permutations from individual diagonal swaps, you can find the types of rotations that perform the entire diagonal permutation at once. Then you not only have a single rotation to apply to the vertices instead of two, but these are also easier to visualise (for me at least) than the diagonal swaps. There are only three types of permutations in $A_4$: the identity, double swaps and $3$-cycles. The corresponding rotations are the identity, rotations through $\pi$ about an axis through opposite face centres, and rotations through $\frac{2\pi}3$ about a principal diagonal.

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  • $\begingroup$ What do you mean by the entire diagonal permutation at once? $\endgroup$ Commented May 24, 2016 at 8:28
  • $\begingroup$ @JacopoStifani: You wrote: "I first apply $(12)(34)$ to the cube, i.e. swap principal diagonals $3$ with $4$ followed by $1$ with $2$, and then record the effect this has on the vertices." I took this to mean that you swap pairs of diagonals twice, one after the other. By "the entire diagonal permutation at once", I mean that you perform the entire operation in one go by finding a single rotation whose effect on the diagonals is the permutation $(12)(34)$, without splitting that up into two swaps. $\endgroup$
    – joriki
    Commented May 24, 2016 at 10:57

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