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I’m jealous; that sounds like a lot of fun. My background is physics, so I’ve never gone through the basic entertaining math, like abstract algebra, but one of these days I’m going to plow through Herstein or something similar until at least I know enough Galois theory to understand why quintics can’t be solved by radicals.

The 3-cycle technique works well for the 3×3 cube but the 4×4 cube can enter states where you cannot solve it using that technique (caveat: at least I can’t see how to leave the state since visual “parity” is preserved by the move). The state I’ve entered with a 4×4 cube is with one edge flipped but its companion edge piece is correctly oriented (all other pieces are correctly oriented).

We talked about that too. It’s solvable by 3-cycles, but it’s hard to see why because the 4 centers of each color are indistinguishable. You can find yourself with two edges in the wrong position, which is impossible to have with a 3×3 cube.

The way to deal with this is to do a center slice (i.e. don’t move the corners but move an internal slice by 90 degrees). This is an odd permutation on the edges and the centers but the centers “don’t care.” If the edges looked odd before (like with a permutation) it’s even now and can be solved with 3-cycles.

Cathy, I don’t know if this would help your students, but I think the story behind your U and V moves is the following assertion. If you take the commutator [g,h] = g h g^{-1}h^{-1} in the symmetric group of two elements which have only one point a in their common support, you get a 3-cycle:(a g(a) h(a)). You might start by asking them to show that if g and h have no common support, the commutator [g,h] is the identity.

Dick,

Cool! That’s a good and more general explanation of why this trick works. I’ll put it in problem set tomorrow night.

Thanks,

Cathy