So we have a cube with N columns, trying to simulate one with n columns. For N=4, we use grouping G={2,2} to simulate n=2, and G={1,2,1} to simulate n=3, with obviously G={1,1,1,1} being the default N=n=4; Where |G|=n. Clearly this is just ways to split up our integer N into different sized sets, subject to certain conditions we hope to derive along the way.
For example, we cannot use 4={1,1,2} as an emulator for n=3, because on the left side we have a face by itself, but we on the right we have the face paired off with an edge section. This means that when solving it, it'll eventually be split up. EG the two (Face, Direction) rotations: (Bottom, Left), (Left Up) would separate part of the 2x2 simulated cube initially in the bottom right corner. This is because of the lack of symmetry in the {1,1,2} expression. IE if we have tiles 'a' and 'b' in the 3rd emulated set (--,--,ab), after a rotation we have (a,b,--), which will then be separated upon moving either column. So we need each tile to be paired with just as many pieces on one side as the other, thus our given emulator distribution MUST be symmetrical.
For all N>3, we we have N={1,(n-2),1}, which is of course symmetrical and has magnitude 3, thus an n=3 cube can always be emulated for any N. For N=5, we also have G={1,3,1}={2,1,2} for n=3 emulation. We cannot have n=2, because we have 5 columns to pair off to our 2 emulated sides. Since we cannot split this evenly, it is not possible. Now in terms of symmetry, it depends on if we have an odd or even number of columns, thus if we have a central column or not.
IE for n odd, we need its central column to be emulated, done differently depending on if N is odd or even. If N is even, we can do this by having equal amounts of columns adjacent to the central divide as being treated as a single block. EG {1,2,1} gives us an emulated central column, so n=3 can be done inside N=4. Note here that the size of that central column (2) is even, but if we have N odd, then the central column in N does not get split up, but instead needs to be bunched with equal amounts of columns on both sides, thus we would have the size of the middle string being odd. EG for n=3 inside N=5, gives G={1,3,1}={2,1,2}, where both middle columns have odd numbers. Hence we have a solution for N odd. Either way, n odd can be emulated in both N even and N odd.
But for n even, we need to split the columns of N into an even amount, one which is clearly impossible if N is odd, thus N also needs to be even. Thus N odd can emulate only n odd, N even can emulate n both odd or even. This is true for All N>n, for example:
N=4. {1,1,1,1}, {1,2,1}, {2,2}.
N=5. {1,1,1,1,1}, {1,3,1}={2,1,2}.
N=6. {1,1,1,1,1,1}, {1,1,2,1,1}, {1,2,2,1}={2,1,1,2}, {1,4,1}={2,2,2}. {3,3}.
N=7. {1,1,1,1,1,1,1}, {1,1,3,1,1}, {1,5,1}={2,3,2}={3,1,3}.
N=8. {1,1,1,1,1,1,1,1}, {1,1,1,2,1,1,1}, {1,1,2,2,1,1}={1,2,1,1,2,1}={2,1,1,1,1,2}, {1,1,4,1,1}={1,2,2,2,1}={2,1,2,1,2}, {1,3,3,1}={3,1,1,3}={2,2,2,2}, {1,6,1}={2,4,2}={3,2,3}, {4,4}.
So Odd cubes can only emulate smaller odd cubes. Even cubes can emulate all smaller cubes. Basically because the emulation will take the symmetric form of either G={---k---,---k'---} or G={--k--,m,--k'--}; thus giving total group size=emulator size of the form a+a=2a=even, or a+b+a=2a+b which can be made even or odd.
