Let's think about a transposition $(3,\ 1).$
It can't be produced by composing $(3,\ 4,\ 1,\ 5),\ (3,\ 6,\ 1,\ 2),\ (6,\ 4,\ 2,\ 5),$ allowing zero or more use of them.
How can we prove that?
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Are you only asking about using those three specific 4-cycles (and possibly their inverses)? In this case, you might see what you can say about the subgroup generated by them. – pjs36 Feb 04 '17 at 01:17
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1@pjs36 Their inverses can be produced by compositing themselves three times? Anyway can we find some patterns of their spanning subgroups? – Ris Feb 04 '17 at 01:28
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How do you know that it can't be produced by composing those cycles? – bof Feb 04 '17 at 02:44
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1@bof In your answer you pointed out that this question is a cube rotation problem. I brought a 3x3 rubix cube. I assumed that the front side's middle tile and back side's middle tile was transposited and then I tried to solve that cube. Solving that cube means that we can transposite arbitary two facing sides' middle tile. That was impossible. So I wondered how to prove that mathematically. – Ris Feb 04 '17 at 02:54
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Sorry for my bad English – Ris Feb 04 '17 at 03:05
1 Answers
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Label the six faces of a cube with numbers $1$ to $6$ as follows:
left $=1,$ top $=2,$ right $=3,$ back $=4,$ front $=5,$ bottom $=6.$
Observe that each of the three given $4$-cycles corresponds to a rotation of the cube, namely, a $90^\circ$ rotation about an axis passing through the centers of two opposite faces. Therefore, the group generated by those three $4$-cycles is a subgroup of the group of rotations of the cube, viewed as face permutations. (In fact it's the whole group, but that's beside the point.) But the transposition $(3,\ 1)$ does not correspond to a rotation of the cube; there is no rotation that interchanges two opposite faces while leaving the other four faces fixed. Therefore, the permutation $(3,\ 1)$ does not belong to the group generated by the permutations $(3,\ 4,\ 1,\ 5),\ (3,\ 6,\ 1,\ 2),\ (6,\ 4,\ 2,\ 5).$
bof
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