Proving impossible Rubik's Cube permutations It turns out that the Rubik's Cube has 12 orbits, because there are a few impossible cases just by turning the edges:

*

*cannot have just one corner twisted (/3)

*cannot have just one edge flipped (/2)

*cannot have only two pieces swapped (/2)

Every document I've read just states these conditions, then goes on to give the number of possible permutations.
How are these conditions arrived at? How has it been proven, for example, that it's not possible to have one edge flipped by just turning?
 A: The essense of the proof is that there’s some invariants in rubik’s cube that legal moves can’t change that invariant, where the configurations you mentioned have different invariants. Similar argument is described in this note about 15 puzzle.
For example, let’s prove the first case. If you have a rubik’s cube with official color arrangements, each corner piece should have exactly one of yellow or white color. Now, let’s ignore other colors and only concentrates on yellow and white on corner blocks. (So it is equivalent to 2 by 2 cube with only two faces colored.) Also, for each corner, let’s assign the number in $\{0,1/3,2/3\}$ depends on how much it twisted compared to the base configuration. (I’ll add pictures later) Then we can prove that all the legal moves (UDFBLR) doesn’t change the sum of the numbers mod 1, i.e. the sum of the eight numbers we assigned for each corner should be integer. However, the case when only one corner twisted in a nontrivial way corresponds to the case where this sum (which is the invariant we found) is not an integer. The other three cases can be proved similarly.
Conversely, it is true that if the sum is an integer, than we can arrange the corners to have 0 twists (not considering their permutations in this case), by using formula that only twists two adjacent corner blocks in a different directions. (This formula is used in blind solving.)
A: I'm gonna try explain how it's impossible to flip just one edge piece, and using the same techniques the other restrictions can be shown too.  This may not work too well without pictures but I'm gonna do my best.
Start by putting an X on one side of every edge piece.  Since there are 12 edge pieces you should have 12 X's total.  Remember the starting location of all your X's.  Now do a basic move, let's start with R.  Now look at the location of all your X's.  Count the number of X's that are NOT in a place where one of your original X's was.  This number should be even!  Now do it with each of the basic moves (RLUDFB) and you'll find it's even every time.
So that proves that no matter how you scramble the cube, you'll always get an even number of X "mismatches".  And flipping just one edge piece will result in an odd number of mismatches.  Therefore, there's no way to scramble the cube that will result in flipping just one edge piece.
