Mathochist:
Actually, if we want a simpler term we may as well just call it the isotropy group.
Omphaloskeptic:
Mathochist:
Anyhow, one thing I’d be careful about is that there aren’t extra positions to get. Your solution is remarkably similar to Jeff Adams’ solution by commutators, but the commutator subgroup of the 3x3x3 magic group is of order 2 (or maybe 4) in the full group, so there might be an extra twist sitting around to be done.
I’m not familiar with Jeff Adams’ solution. The group generated by the “primitive moves” (inner and outer slice rotations) leaves the centers in fixed positions, so its commutator subgroup will have trivial center rotations, right (each inverse move undoes the original move’s center rotations, since the rotation groups are abelian)? What moves does he consider, to get nontrivial commutator subgroups?
There’s a poster around with his solution. I learned it from himself, so…
Anyhow, your solution seems to also be by commutators, which is why I brought it up. Basically, if the group of moves is G, then the moves you’re talking about using to act on the set of inside corners and inside edges can only generate [G,G]. Now, if your full solution uses a different technique, that changes things…
Ah, I misread your earlier note. I read it as saying that the commutator subgroup of the isotropy group (instead of the magic group) had order 2, which confused me because the isotropy group is abelian. What you actually said makes more sense.
Omphaloskeptic:
Ah, I misread your earlier note. I read it as saying that the commutator subgroup of the isotropy group (instead of the magic group) had order 2, which confused me because the isotropy group is abelian. What you actually said makes more sense.
Yeah, I suppose I did move from asking about your solution’s application to the isotropy group to asking about the solution in general without properly signaling the cognitive lane change.