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The corner group of 4^3 is exactly like that of the 3^3. It has 8!*3^7 elements.

Now to calculate the order of the whole group, we have to find

the order of the corner stabilizer group. That's the group of

moves that leave the corners fixed.

SO let's start by thinking about the edge group of the corner

stabilizer. I personally have a tool that exchanges two edge cubies

without moving corners. Since the edge group is transitive, I can

exchange any two edges without moving corners.

You cannot flip any edge of the 4^3 without moving it. So there's no

edge-flippage in the 3^3 sense on the 4^3. That means that the edge

group in the corner stabilizer has order 24!.

Now all there is left is compute the order of the edge AND corner

stabilizer. The inside twists (slices) are center-even. The outside

twists (faces) are center-odd, but they are also corner-odd, and so if

we want to bring the corners home we have to make an even number of

outside twists. That means that all the center-permutations in the

edge-and-corner stabilizer are even. But how many of these can be

achieved?

I have a tool that exchanges two pairs of centers. I think (but I

haven't yet proved) that the center group is 4-transitive, so that

ANY two pairs of centers can be exchanged. That means that all the

even permutations of centers can be done without perturbing edges and

corners. Hence the size of the edge-and-corner stabilizer is 24!/2.

As for supergroupiness, Dave Plummer just pointed out to me that

once you know a center's position, you know its orientation, since

center cubies always keep one corner pointing to the middle of the

face. So 4^3 has no supergroup.

So the order of the 4^3 group is 8!*3^7*24!*24!/2. In numbers, if you insist, |4^3|= 16,972,688,908,618,238,933,770,849,245,964,147,960,401,887,232,000,000,000 or about 1.7*10^55.

Now that number is a little deceptive, because it includes whole-cube

rotations. The 4^3 has no nice fixed reference frame like the 3^3 has.

If you don't want to count whole-cube rotations you have to divide by

24, to get

707,195,371,192,426,622,240,452,051,915,172,831,683,411,968,000,000,000 or about 7.1*10^53.

Finally, we have to face the fact that the center cubies are

in six nonintradistinguishable sets of four. (All the edge cubies are

distinguishable by color or orientation). So we have to divide our

last result by 4!^6/2. That leaves

7,401,196,841,564,901,869,874,093,974,498,574,336,000,000,000

or about 7.4*10^45 distinguishable color patterns. Remember that these

do not form a group.

I leave it as an exercise for Hoey and Saxe to find a lower bound for

the diameter of the group.

---Wechsler