The Subgroup Q and its cosets 
If we solve all 138,639,780 essentially different Hcosets, we still solve almost all possible cube positions three times. This is because we are able to reduce the number of Hcosets only by D4h symmetries (recall that there are 16 of them), while the symmetry group Oh of the cube has 48 elements. Let's observe what happens if for example we rotate the coset H*R, about the long diagonal containing the corners URF and DLB, by 120 degrees.

H*R =
Rotating this color pattern about the URFDLB axis we get
and the result after exchanging the colors again so that the colors for the centers are restored is
This does not look like a coset of H any more and indeed it is not! Look at the corners first. Because the white facelets have vanished, the orientations of the corners are now undefined. The positions of the redgreen edges have also vanished.
We need to relabel the cube in a different way to be able to retrieve information about Hcoset properties when doing some symmetry operations which involve rotations along the long diagonal. 
We accomplish this by coloring additional facelets in the Hsubgroup color scheme. We let Q be the subgroup corresponding to this new color scheme, and we also denote the corresponding color pattern by Q.
Pattern Q = 
Obviously subgroup Q is a proper subgroup of H. The moves U2, D2, R2, F2, L2, B2 do not change the Qwise relabeled cube. In analogy to the subgroup H one could argue that <U2, D2, R2, F2, L2, B2> generates Q, but this is not true. In this way way you only get a subgroup of index 6 in Q.
Observe that there a two different types of corners, one with the colors white, green and red in clockwise orientation and one with these colors anticlockwise. The positions of each of these types build a tetrahedron, so there are 4!*4! possibilities to exchange the corners without changing the relabeled cube. Analogously there are 4! ways to exchange the redgreen edges, 4! ways to exchange the whitegreen edges, and 4! ways to exchange the whitered edges. Because we can generate only even permutations without disassembling the cube, the subgroup Q has (4!)^5 / 2 = 3,981,312 elements. The index of Q in H is 19,508,428,800 / 3,981,312 = 4900.
It is easy to spot if a cube position is an element of Q. All faces only have only two colors, the center facelet color or the color of the opposite center facelet. Here is an example:

The subgroup Q has all 48 symmetries of the cube. This is why we usually find 48 cosets which are equivalent to each other.
Q*R = , Q*F = , Q*L=,
Q*B = , Q*U = and Q*D =
for example are all equivalent to each other. In this case there are only 6 equivalent cases, because this coset has 8 selfsymmetries.

The number of cosets of Q is
G/Q = 43,252,003,274,489,856,000 / 3,981,312 = 10,863,756,288,000, but we can get a better understanding if we compute this number directly by counting the number of possible color patterns of the relabeled cube.
There are C(12,4) possibilities to set the redgreen edges and C(8,4) possibilities to set the whitegreen edges, and then the whitered edges are forced. There are C(8,4) ways to set the 4 whitegreenred anticlockwise oriented corners, and the 4 whitegreenred clockwise oriented corners are forced. The corners may be oriented in 3^7 ways and the edges in 2^11 ways. So the total number of cosets of Q is C(12,4) * C(8,4)^2 * 3^7 * 2^11 = 10,863,756,288,000.

When we reduce the number 10,863,756,288,000 of Qcosets by the 48 symmetries of Oh it can be shown that there are 226,331,259,372 essentially different equivalence classes. This is about 10,863,756,288,000 / 48, but not exactly. It is a bit more because some equivalence classes have less than 48 elements as explained before.

Because Q is a subgroup with index 4900 in H, each Hcoset splits into 4900 Qcosets.
Take for example the Hcoset
H*R =
If we complete the Q color scheme there are C(8,4) = 70 ways to set the 4 anticlockwise whitegreenred corners, for example:
Now there are C(8,4) = 70 ways to set the whitegreen edges for example:
At this point the 4 whitered edges and the 4 whitegreenred clockwise corners are forced. The result is an example of one of the 70*70 Qcosets which build the H*Rcoset.

If we were able to select a subset of the 138,639,780 essentially different Hcosets containing 4900 Qcosets each so that each of the 226,331,259,372 essentially different Qcosets (or a coset equivalent with reference to Ohsymmetry) is contained in at least one element of the selected Hcosets, the entire cube space could be essentially solved by solving only this subset of the Hcosets.
This is a typical set cover problem, aside from the fact that the number 138,639,780 of involved sets is too large to find a good solution using available techniques and hardware. We can get around this obstacle if we build unions of Hcosets and deal with a significantly smaller number of sets.
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