Lemma 5:
I claim that the parity of the corner orbit and the parity of the innermost edge orbit must be the same, and therefore either both are even or both are odd.

Proof:
In order to prove this we have to find, if it exists, all the possible moves that affect the parity of the edge and corner orbits alone, and all the ones that affect both, and any possible combinations of the two types. We can see that the only moves that affect the permutation the corners are the 18 outer face moves, and each of these moves also affects the permutation of the innermost edges. Also remember that we have banned the turning on an innermost slice as an allowable move, which would be the only other move that would affect either of these types of pieces.

Therefore we only have to examine the 18 outer face moves, and how they affect the corners and innermost edges. We can break these moves into two types. The first type are the 12 quarter turn moves F, F', B, B', R, R', L, L', U, U', D, D'. The second type are the 6 double turn moves F2, B2, R2, L2, U2, D2.

1) First we will examine the quarter turns and how they affect the permutations of the corners and edges. Start from the solved position of the cube and perform any of the 12 outer layer quarter turns. We can see that the quarter turn 4-cycles the four corners in the layer as well as the four innermost edges in that layer. A four-cycle is in affect doing three transpositions, which makes the permutation of the corners and edges now odd instead of even. Given any permutations of the corners and edges (we do not yet know whether they are both the same or both different), then a quarter turn will change the parity of the corner orbit and the edge orbit.

2) Next we will examine the double turns and how they affect the permutations of the corners and edges. Start from the solved positions of the cube and perform any of the 6 outer layer double turns. We can see that this turn will perform two transpositions on the corners and two tranpositions on the innermost edges. This turn performs and even number of transpositions, and hence it does not affect the parity of either the corner or the innermost edge orbit. Therefore an outer layer double turn does not affect the parity of either the corner or innermost edge orbit.

So we can see that an outer layer quarter turn will change the parity of both the corner and edge orbits, and that a double turn does not affect either orbit. Since we know that the solved state must be an allowable position, and in the solved state both the corner and innermost edge orbits have even parity, then the only way to affect the parity of either orbit is to perform an outer face quarter turn and change both parities.

Therefore we can conclude that the parity of the corner and innermost edge orbits must be the same, and that either both are even or both are odd and the lemma is proved.

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