Arie Groeneveld wrote:
>
> What I learned so far from group theory is:
> The size (order) of the group of permutations in this case is
> !6
> 720
> The size of the alternating group is
> -: !6
> 360
> and that's equal to
> # P
> 360
> in Roger Hui's solution and,
>
> from Raul's remarks
> p=: o (] ~.@, ,/@:({"1/))^:_,:n
> # p
>
If you want to prove that the group H of the 1D Rubik's cube is in fact
A6, without enumeration, it suffices to show that H is a normal subgroup
of A6, since A6 is simple. If you cannot see a quick way of doing this,
you are reduced to calculating orders of stabilizers. H acts transitively
on single letters, so the 1-letter stabilizers are isomorphic and have
order |H|/6. In particular 6 divides the order of H. You then work your
way up. If you can prove that H is 4-transitive, you are immediately
done.
Best wishes,
John
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