Want to find the orbits of the group of $24$ rotations of the cube.
The rotations about the axis, formed by the midpoints of the anti-podal faces parallel to the $x-z$ plane have the $3$ non-trivial symmetries as:
- $\alpha=(1234)(5678),$
- $\alpha^2=(13)(24)(57)(68),$
- $\alpha^3=(1432)(5876),$
And, the symmetries along the axis, formed by the midpoints of the anti-podal faces parallel to the $x-y$ plane have the $3$ non-trivial symmetries as:
4: $\alpha'=(3762)(4851),$
5: $\alpha'^2= (36)(27)(81)(45),$
6: $\alpha'^3= (3267)(4158),$
If want to compute $\alpha\alpha'=(1234)(5678)(3267)(4851),$ then have two doubts.
Is composition of symmetries along perpendicular axis, meaningful in practice?
Get two maps for many elements. Say, for the element $2:$
$2\rightarrow 7, $ and $2\rightarrow 3;$ as shown below in (f), (n) respectively:
Processing permutations from right to left, with individual map being processed from right to left:
First permutation from the right:
(a). $4\rightarrow 8, 8\rightarrow 5 : 4\rightarrow 5,$
(b). $8\rightarrow 5, 5\rightarrow 6: 8\rightarrow 6,$
(c). $5\rightarrow 1, 1\rightarrow 2: 5\rightarrow 2,$
(d). $1\rightarrow 4, 4\rightarrow 1: 1\rightarrow 1,$
Second permutation from the right:
(e). $3\rightarrow 2, 2\rightarrow 3: 3\rightarrow 3,$
(f). $2\rightarrow 6, 6\rightarrow 7: 2\rightarrow 7,$
(g). $6\rightarrow 7, 7\rightarrow 8: 6\rightarrow 8,$
(h). $7\rightarrow 3, 3\rightarrow 4: 7\rightarrow 4.$
Third permutation from the right:
(i) $5\rightarrow 6,$ get second map for the vertex $5,$
(j) $6\rightarrow 7,$ get second map for the vertex $6,$
(k) $7\rightarrow 8,$ get second map for the vertex $7,$
(l) $8\rightarrow 5.$
Fourth permutation from the right:
(m) $1\rightarrow 2,$ get second map for the vertex $1,$
(n) $2\rightarrow 3,$ get second map for the vertex $2,$
(o) $3\rightarrow 4,$ get second map for the vertex $3,$
(p) $4\rightarrow 1,$ get second map for the vertex $4.$
What should be the final map is unclear.
It cannot be \begin{pmatrix} 1&2&3&4&5&6&7&8\\ 2&3&4&1&6&7&8&5 \end{pmatrix} $= (1234)(5678)?$
as that is illogical, as the second rotation made no effect.
Hence, should such (of the symmetries along the perpendicular axis) composition be ignored simply?
If so, how to fill the corresponding entries in the group table?
Edit : The permutations are a bijective map. Hence, the composition of permutations is to be taken for any element in any permutation, once only.
The main error above is that the last two permutations, i.e.$(1234)(5678)$ are taken again. Also, there is error in taking one permutation of $\alpha'^3$ instead of $\alpha.$
The same is being pointed by the first comment of @JyrkiLahtonen.