That's a very complex post, interesting, good explanation. Maybe this isn't exactly in 'group theory' but can you explain how having say a single edge flipped (by taking a single piece out and flipping it), or some other such 'impossible' permutation/orientation (on a 3x3) violates the concepts you've explained?
It does not exactly violate the concepts of group theory. Maybe I should make it a bit clearer what I mean by a "generated subgroup". I said that the group generated by a set of elements is the smallest group containing all these elements. This is not very intuitive; a better way to look at it is to say that this group contains all elements that can be obtained by multiplying the generators and their powers and inverses. So, the group generated by U and R also contains R'UR. It does not however contain, say, D.
To distinguish a set of elements from the group it generates I use curly braces for sets, and angle brackets for generated groups. So the group generated by {U,R} is denoted by <U,R>. Hence, the whole group of the cube is <U,R,L,D,F,B>.
Note that for a given group there is no unique set of generators. You could also write the same group as, say <U,R'U'R,L,D',F>.
Now the "parity" problem of a single flipped edge that you referred to simply states that the permutation that flips a single edge is not contained in this group. In other words, it cannot be written as a product of the original generators. So if we face such a situation we can't solve it legally (i.e., without disassembling the cube).
I don't know a proof of this fact right now, but my guess is that it's a bit more involved. Something along the lines of defining orientation for edges even if they are not permuted correctly, and then showing that each of the generators changes the orientation of an even number of edges. Thus you can never arrive at a situation where an odd number of edges is oriented incorrectly.
Also, this is more just something I'm wondering than something I want an answer to, but I'm wondering what can be said about symmetry between groups, mainly how often they occur (if at all) and how applicable they are to solving (say, with HTA?). Is symmetry the right word? Maybe you get what I'm saying, a position which belongs to two distinct groups (are groups mutually exclusive?)
Groups are certainly not mutually exclusive. Take the group <R2>, which has two elements, namely {id, R2}, where id is the neutral element, i.e., the permutation that doesn't change anything. Now look at the group <R> which contains the elements {id, R, R2, R'}. We see that these two groups share two elements, id and R2. In fact, <R2> is fully contained in <R>, hence we call it a subgroup of <R>.
One interesting fact to note is that the
intersection of two groups, that is, the collection of elements which belong to both, forms again a group.
Not sure if I answered your question. But certainly a given configuration can belong to several groups.
EDIT: It seems strange to me that a position that is in {U2,D2,B2,F2,L2,R2} could also belong (I think) to {U,D,B,F,L,R} and maybe also any other group. When you say it belongs to the former, are you assuming that it belongs to all sub-groups within it, or that those are irrelevant, or what?
If you change the notation to <U2,D2,B2,F2,L2,R2> and <U,D,B,F,L,R>, then yes, an element can belong to both of them. In fact, since the first is a subgroup of the second, any element in the first group also belongs to the second one.
For the second part you have it backwards. If an element is contained in one group, then it is also contained in all overgroups (which is just the opposite of subgroups).
Second stupid question edit: If a cube is in that former group (all 2 turns), could it be solved using only commutators that also fit into that group? I'm going to be trying this out...
This is a property specific to the group in question, and I don't know the answer (but I'm going to find out). For the whole group <U,D,B,F,L,R> it is true, at least if you use "commutator" in the mathematical sense. So, any scramble can be written in the form xyx'y' for an appropriate choice of algorithms x and y.
Edit: Actually what I wrote in the previous paragraph is incorrect. What is correct is that any scramble can be solved using a sequence of commutators. For the half turn subgroup this is not the case. For example, U2 can't be solved using commutators made up of half moves.
