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I found these two papers online which concurs with the 765765 being the highest order for the 4x4x4. The first is in German, but the 2nd one appears to be very much related and in English.

These papers give the following order-765765 maneuver: U Lw b' x'

(I've translated their expression into WCA notation.)

The maximum order for an element of the <U,R,F> 2x2x2 group is 36. (qqwref's value of 45 for the <U,D,L,R,F,B> 2x2x2 group is correct for that group, of course.)

I note that in general, when talking about orders of maneuvers of various puzzles, one needs to be careful in stating exactly what you mean. Not only is there cube rotations vs. no cube rotations (as in <U,D,L,R,F,B> vs. <U,x,y,z> groups), but also whether or not rotating of a central face piece matters or not. I also note that many cubes sites will define "order" differently from the mathematical definition.

The mathematical term defines the order of a group element as the smallest positive power the group element must be raised to so that the result is the identity element of the group. On cubing sites, the order of a maneuver is often defined as the smallest number of times the maneuver must be performed (starting from a solved state) to make the cube solved again. These definitions are not equivalent!

First of all, the standard 3x3x3 "cube group" is commonly considered to have 43,252,003,274,489,856,000 elements and can be denoted in generator notation by <U,D,L,R,F,B>. In this group, the center pieces are considered fixed reference pieces with respect to which we interpret the various moves (U, D, etc.). It is often desirable (especially in the context of specifying speedsolving maneuvers) to not consider the center pieces to be fixed references, but to think of the cube as having an external frame of reference. Since the center pieces can now be moved around in 24 different arrangements with respect to this reference frame, cubing notation became extended with new symbols such as x, y, z to represent 90-degree whole cube rotations. We can describe this group in generator notation by <U,x,y,z>. (We could also say <U,D,L,R,F,B,x,y,z>, <U,D,L,R,F,B,M,E,S> or even just <U, x, y>, as long as the generators listed can be combined in some way to produce every possible transformation allowed by this group.)

The <U,x,y,z> group is 24 times larger than the standard cube group. Note also that if we start from the solved state, the "move" x does not change the cube from being solved to being not solved. It only changes the orientation of the cube with respect to the reference frame. So while x does not change the "intrinsic state" of the cube, it does make an important change with respect to the reference frame, so it is not an identity element, but an element of order 4. Doing x four times always brings the cube back to the state (with respect to the reference frame) that we started with. So the first way that the definitions of "order" above can be different is in the fact that the group may represent several "solved states" simply by representing a cube in 24 different orientations, whereas there is never more than one identity element in a group.

The 4x4x4 and larger cubes generally have pieces on each face that are considered identical. So a maneuver (applied some number of times) might move such identical pieces around within the same faces, and thus would still be considered solved. However, if we apply the maneuver to some other cube state, we may see that there is an effect on the pieces. But in a group, only the identity element has a null effect. The set of "positions" of the puzzle is not compatible with the mathematical concept of a group. However, the way the actual pieces can be rearranged by maneuvers can still be modeled as a group (for standard, non-bandaged NxNxN cube puzzles, anyway). But the identity element of such a group is the "null transformation" that doesn't move any piece somewhere else. So the mathematical definition of "order" would mean you apply the maneuver until every piece is back to its original place, even if a smaller number of applications of the maneuver might make the puzzle looked solved (and even in the same cube orientation) because of identical pieces being permuted among themselves.

Anyway, we can also talk about orders of elements for the supercube group. For the <U,D,L,R,F,B> 3x3x3 supercube group, the maximum order of an element is 1980. I'm not aware if the maximum order of the <U,x,y,z> 3x3x3 supercube group has been reported before. I am fairly convinced it is 5040. An element having that order is:

R L2 U S2 D L2 U L' E2 L U Rw2 D2 R2 D' L' E2 S L S Lw E' L' E' S2

If 5040 is not high enough for you, we can consider the illegal 3x3x3 supercube group (where disassembly reassembly is allowed, but not restickering) with cube rotations allowed. I'm pretty sure the highest order in that group is 7920.

On cubing sites, the order of a maneuver is often defined as the smallest number of times the maneuver must be performed (starting from a solved state) to make the cube solved again. These definitions are not equivalent!

Right, they're not. I like the "until it's solved" definition more, though, because it matches the way you'd play around with such sequences on a real cube - by doing them until the cube is solved again.

For the <U,D,L,R,F,B> 3x3x3 supercube group, the maximum order of an element is 1980. I'm not aware if the maximum order of the <U,x,y,z> 3x3x3 supercube group has been reported before. I am fairly convinced it is 5040. An element having that order is:
R L2 U S2 D L2 U L' E2 L U Rw2 D2 R2 D' L' E2 S L S Lw E' L' E' S2

If 5040 is not high enough for you, we can consider the illegal 3x3x3 supercube group (where disassembly reassembly is allowed, but not restickering) with cube rotations allowed. I'm pretty sure the highest order in that group is 7920.

For the <U,D,L,R,F,B> 3x3x3 supercube group, the maximum order of an element is 1980. I'm not aware if the maximum order of the <U,x,y,z> 3x3x3 supercube group has been reported before. I am fairly convinced it is 5040. An element having that order is:

R L2 U S2 D L2 U L' E2 L U Rw2 D2 R2 D' L' E2 S L S Lw E' L' E' S2

If 5040 is not high enough for you, we can consider the illegal 3x3x3 supercube group (where disassembly reassembly is allowed, but not restickering) with cube rotations allowed. I'm pretty sure the highest order in that group is 7920.

3x3x3 supercube group (without rotations)
1980 = 11 * 45 * 4.
That's 11 for the edges, 45 for the corners, and 4 for the centers.

3x3x3 supercube group (with rotations)
5040 = 7 * 45 * 16.
That's 7 for the edges, 45 for the corners, and 16 for the centers.
The centers group of the 3x3x3 supercube (with rotations) has elements of order 1, 2, 3, 4, 6, 8, 12, and 16. Order 16 requires transformations having void cube parity. Since you need an even permutation of the corners to get the 45, you need an odd permutation of edges to satisfy void cube parity. Hence, you can't use a factor of 11 from the edges since you also need a 2-cycle or a 4-cycle.

3x3x3 illegal supercube group (with rotations)
7920 = 11 * 45 * 16.
With the illegal supercube, parity constraints are removed.

Nope. U colors are too hard to find in some block combinations for this to work for that. The non-matching version, which I've been working on along with the other NMLL on my site, is to orient the L/R colors to U and separate the edges. That is 38 cases but has more difficult recognition than if L/R colors were separated as in the original.

One way I've thought to know the 3-cycle is to, while finishing 4b, watch the edge that goes to DF. With M'U2M' for example it's easy to see that the UB sticker will be at DF. Then you'll see where its friend is on U.

- If the U friend matches the U center, place it at UF, M'U2
- If the U friend doesn't match the U center, place it at UB, MU2

What it doesn't work well for are the M2U2M(U2) and M2U2M'(U2) cases. I like this better than the L/R same/opposite colors way that a couple of people have talked about in the Roux guide topic, because you don't have to worry about whether you are supposed to align the corners correctly or align the edges according to the F/B centers.

With this way, you don't have to think about the F/B centers or the UL/UR edges.

Place the U friend at UB (whether it matches the U center or not). And if you instead watched the edge that went to DB, place it's U friend at UF. I've done a bunch of solves and this is very easy. And for the M2U2M(U2) and M2U2M'(U2) cases, it only adds one more move.

Examples (scramble all with orange on U, green on F):

U M U2 M U M2 U' M U M2 U M' U' M' U M U M2 U2 M2 U M2 U M U' M'

Orient + place - M2UM'UMU2M2
With this, you should have seen that during, and a little before, the last M2, A red edge went to DF. And you see that the other red edge is at UR. So place it at UB and do MU2 (because it doesn't match the center).

M2 U2 M' U2 M2 U2 M U' M2 U M2 U2 M U' M U M U' M U2 M U' M2 U M'

Orient + place - M2U'M'UMU2M2
During U2M2, you see that a red edge went to DF. And you see that the other red edge is at UR. Place it at UB and do M'U2 (it matches the center).

U M' U' M' U M' U M' U2 M2 U' M U M U M2 U M U M2 U2 M' U M2 U'

Orient + place - UMUM'UM2U'M'U2M'
During M'U2M', you see that an orange edge went to DF. And you see that the other orange edge is at UR. Place it at UB and do MU2 (doesn't match the center).

This is so easy. Sometimes you'll instead watch an edge that goes to DB, and it works the same way. There is no recognition time for this as with the "rotate cube", "tilt cube", or this. It's all lookahead and there is no thinking time. After only a few solves, the lookahead for this became a habit.