we just talked about this in university and i was curious about normal subgroups on 3x3.
i found a few obvious ones so far:
identity, whole group (haha)
edge flips, corner twists
edge permutations, corner permutations
(and of course you can link the ones that are disjunct and get a new one.)
are there any more? if no, is there a proof for this?
what about puzzles with only one type of pieces and fixed orientation (e.g. dino cube). do they have normal subgroups besides the identity and the whole group?
to see a 7.19 OH solve
There's the centre of the group (on the cube that is the group consisting of the identity and the superflip).
There's the commutator subgroup (which on the cube consists of all even permutations incl. twists and flips).
There's the group of even permutations (without twists and/or flips).
On most puzzles you can just look at each orbit (set of interchangeable pieces) individually, and them note its permutation normal subgroups (which is usually just the even permutations) and its orientation normal subgroups (which usually flips/twists all the pieces the same way). Combining these then gives all normal subgroups of the whole group.
As for the Dino cube, keeping one edge as a fixed reference, it has group A11 (group of even permutations of 11 things). The groups An with n>=5 are simple, i.e. have no non-trivial normal groups. So the Dino cube group has none either.
For a more interesting exercise, try finding all normal groups of the Floppy cube.
Last edited by jaap; 01-21-2012 at 02:15 AM.
Except that this one is not a group. The group generated by the even permutations is in fact the full commutator subgroup.
Originally Posted by jaap
For example, (UB -> UL -> UF -> UB) combined with (UB -> UF -> LU -> UB) yields a 2-edge flip (UF and UL).