# Inversion by conjugation

#### whauk

##### Member
the question i will discuss here is the following:
for a (random) given permutation A, does there exist a permutation B with B A B'=A' (which is equivalent to B A B' A= id)? or in other words: can you invert any permutation by conjugation?
i start by discussing the problem in the symmetric group and will try to expand the results to the cube group. so for understanding this article it might be helpful to know a little bit about group theory and also cycle notation will be used to describe permutations. and as always: id = () = identity = cycle of nothing and € means "element of a set".

motivation (this can be skipped without missing relevant information):

as we all know R U R' U R U2 R' L' U' L U L' U2 L twists two corners on the cube. as linus discovered this one R U2 R' L' U' L U' R U' R' L' U2 L U also twists two corners but is not reached by setup-moves from the original algorithm. however it seems to be very similar and becomes a conjugate of the original alg if you replace the bold part by L' U2 L R U R' which does the same thing. so: R U' R' L' U2 L=L' U2 L R U R'. with A=R U' R' and B=L' U2 L the equality simplifies to AB=B'A' and this is equivalent to BAB'=A'. so it might be possible to simplify/modify algorithms if you find a B for A such that BAB'=A'. this might not work very often but the question became so interesting that i started thinking about it seriously.

theoretical approach:

any permutation can be written as a disjunct cycle of stickers. we will consider those cycles as elements of a symmetric group. so lets look at one of these cycles, namely (1 2 ... n):=A first:
there is an obvious B with the property BAB'=A' because the "mirror" of the cycle does what we want. so set M as M:=(1 n) (2 n-1) (3 n-2).... in general M(i)=n+1-i for every i between 1 and n. it is very easy to verify that this works so i don't feel like writing it down. probably this image helps you to understand how it works:

take a polygon with n corners labeled 1 ... n in clockwise direction. the cycle (1 2 ... n) represented on the polygon is a rotation by one corner in clockwise direction. if you mirror the polygon first ("flipping the paper it is drawn on") rotate it clockwise and flip the paper over again it will result in being turned counterclockwise which occurs to be the inverse of turning to the right.

to keep it as general as possible we notice that if C and D are in the commutator group of A (the subgroup of all elements that commute with A (=> also with A')) CMD:=B has the same property BAB'=A' since BAB' = CMDA(CMD)' = CMDAD'M'C' = CMDD'AM'C' = CMAM'C' = CA'C' = CC'A' = A'.
remark: i am not sure whether there are even more elements in the group with the desired property. i couldn't find any and tried to prove that there are none but i didn't succeed. if you have an answer let me know
also the set {B | B = CMD for M is mirror of A and C,D commute with A} seems very strange to me. its like the left coset of a right coset?! did you ever encounter such a structure? does it have a name or any cool properties?

for now we have only looked at one cycle. however if you are confronted with a permutation consisting of the cycles A1,...,Am you can just find Bi for every Ai (that doesn't disturb Aj with i≠j) and B1...Bm will do the job.

applying the theoretical results on the cube (edges):

stickers on the same piece always stay on the same piece so they actually do not move around freely. (therefore a possible B might actually not exist with the restriction that stickers must stay on their piece) the funny thing is that our theoretical works perfectly out for arbitrary edge permutations.
if the edge-cycle is "pure" (the sticker-cycle does not hit a piece twice. so no edge in the cycle is additionally "flipped") we can replace "stickers" with "pieces" and the restriction does not longer bother us. (example: A= (UR UF UL), => B=S2 D2 S2 which is just the mirror)

if not (the cycle is not pure) things get a little more complicated but it still works:
so k edges are cycled and the sticker cycle can be written as (1 2 ... 2k):=A. i hope this makes sense to everyone. at k the cycle reaches its original edge but on the "other sticker", then proceeds on the "other stickers" of each edge and reaches its original position at 2k. therefore two stickers are on the same edge if and only if their distance is k in the cycle which also defines our restriction.
the mirroring M applied on stickers gives: M(i)=2k+1-i and M(i+k)=2k+1-i-k=k+1-i. => M(i)-M(i+k)=2k+1-i-(k+1-i)=k which means that stickers on the same edge (they have difference k) still have difference k after applying M. so in fact M exists as a permutation of edge pieces (before we only knew it existed as a permutation of stickers which might not be doable on a cube).

and on corners:
this very satisfying conclusion does not work with non-pure corner permutations. (you would have to replace 2k with 3k and everything gets weird).
i actually found a proof that for A given by U' L' U F2 U' L U2 L' F2 L F2 U' F2 (basically a corner 3-cycle and two twisted corners, one in the 3-cycle, the other one not. this was the easiest example of what doesn't work with the standard method. (later i found that a 3-corner twist is a much more obvious counterexample. however i still think its interesting to know that this one doesn't work either)) no B with the desired property exists. i will only provide a sketch as i do not feel writing it all out correctly:
lets first assume B exists. there are two cases:
1. the solely twisted corner (DFL) is not moved into the 3-cycle by B (or B'). the corner must be solved again by BAB'A. but A doesn't affect the corner's permutation => the permutation of our corner after BAB'A is already given by BB'=id and this also means that the corner orientation is given only by AA. we know that A twists the corner once and two corner twists will not solve it again. => BAB'A is not id in this case.

2. B makes the corner (DFL) interact with corners in the 3-cycle. in this case consider only the corner permutation. A can then be written as (1 2 3) and B is a permutation that has 1 and/or 2 and/or 3 and sth else in the same cycle.
i will now look on what is happening in detail when applying BAB'A to the cube which should be the identity.
without loss of generality lets assume that applying B makes the DFL corner go to 1 (an arbitrary corner in the 3-cycle that i name 1). applying A then makes this corner go to 2 and after another B' it has to go back to DFL because the permutation of this corner is A-invariant. so this means B(DFL)=1 and B'(2)=DFL. this is equivalent to B(DFL)=1 and B(DFL)=2 which cannot be possible.

as we have covered all possibilities for B and none of them has the desired property of BAB'A=id we know that B cannot exist. therefore the permutation given by U' L' U F2 U' L U2 L' F2 L F2 U' F2 is indeed one that cannot by inverted by conjugation.

Conclusion:

i believe that any permutation of the cube that has only pure corner cycles and as many corners twisted in spot clockwise as counterclockwise can be inverted by conjugation. (heuristic explanation: two corners twisted in different directions need a swap of themselves as setup move (so it works). furthermore i think the length and number of unpure corner cycles doesn't change much from the above counterexample and it still doesn't work. aaaaand finally i think in the case A where you would need a setup move M that violates the permutation laws of the cube you can consider the group <R,U,L,F,B,D,two-edge-swap> and you will find "enough" elements C,D there such that a B€{B | B = CMD for M is mirror of A and C,D commute with A} lies in the cube group.) or in other words: i think you do not get problems because of the fact that you cannot swap only two pieces on a cube.
this is what i guess is true! finding a proof for everything i claimed in the last paragraph would probably result in disastrous case differentiations and the only people you would like to read it might be your mortal enemies. if you find something out let me know about your results.

i hope you found the question (and answer) an interesting-to-discuss problem and hope you understand everything (as i tried to make it as easy to read as possible). however if anything is unclear, dont hesitate to ask me

edit:
TLDR version: the answer is no.

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#### Forte

##### Member
Cool idea! I'll try to get around to looking into it later, kinda busy now ):

remark: i am not sure whether there are even more elements in the group with the desired property. i couldn't find any and tried to prove that there are none but i didn't succeed. if you have an answer let me know
Let $$S_n$$ be the symmetric group on $$n$$ elements (but I'm going to view it as the set of bijections on $$\mathbb{Z}/n\mathbb{Z}$$).
Let $$p \in S_n$$ be defined by $$p(n) = n+1$$ (this is just that $$n$$-cycle).
Also instead of $$bab^{-1} = a^{-1}$$, I'm going to use $$aba = b$$ so there's less writing lol

Claim: $$\left\vert\{f \in S_n \mid pfp = f\}\right\vert = n$$

Proof:
Note the following equivalences:
$$pfp = f$$
$$\iff (pfp)(n) = f(n) \forall n$$
$$\iff (pf)(n+1) = f(n) \forall n$$
$$\iff f(n+1) + 1 = f(n) \forall n$$

First of all, note that this condition means that each function $$f$$ satisfying this property is determined by $$f(0)$$ (because the last line will give you all values of the function), so there can be at most $$n$$ distinct maps.
We then note that the family of maps $$g_i(n) = -n + i$$ (for any $$i \in \mathbb{Z}/n\mathbb{Z}$$) satisfy this condition since:
$$g_i(n+1) + 1 = -(n+1) + i + 1 = -n - 1 + i + 1 = -n + i = g_i(n)$$.
We've found $$n$$ distinct maps, so we're done!

Also didn't there used to be latex tags here? (EDIT: Thanks Stefan )

Oooooh also turns out there's a name for this: Real element of a group

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#### Stefan

##### Member
Also didn't there used to be latex tags here?
Try $$. Moritz, would be nice if you started your sentences with capital letters.$$

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#### Christopher Mowla

the question i will discuss here is the following:
for a (random) given permutation A, does there exist a permutation B with B A B'=A' (which is equivalent to B A B' A= id)? or in other words: can you invert any permutation by conjugation?
When I first saw this thread, I was fascinated with this question, and today, for the first time since this thread has been posted, I felt that I might take a shot at finding the answer, and I have. I apologize for the delay (and I'm sorry to see that no one seemed to be interested to find an answer after more than a year).

I just created a method (algorithm) to find the inverse by conjugation, and from the method, I have drawn the following conclusion.

My Conjecture
For all 6 colored 3x3x3 Rubik's cube positions A in which their order is equal to the least common multiple of all their disjoint cycles of corners and middle edges, there exists multiple positions B such that B A B' = A'. However, for all 3x3x3 Rubik's cube positions A in which the order of the position is NOT equal to (larger than) this least common multiple, there are no positions B such that B A B' = A'.

Therefore, the position A generated by the maneuver, R U R' U R U2 R' L' U' L U L' U2 L has multiple positions B such that B A B' = A'.

Example
I grabbed this random odd permutation scramble from qqtimer.
Let A = U' B2 D' R2 D2 B2 F2 R2 U' R2 D B D2 F L2 U R' D' U' R2 B'

This is one such position which, according to my conjecture, does NOT have any sequence B such that B A B' = A'.

However, I then modified this scramble (I twisted two more corners which were originally oriented correctly) to become a position which we can find such a sequence B for.

Let A = U' B2 D' R2 D2 B2 F2 R2 U' R2 D B D2 F L2 U R' D' U' R2 B' [ F', D' B' D][ D, L' U' L] = D' B F' D' L2 B' R D L2 U L U L' R F2 U2 F2 U',
then one such position B such that B A B' = A' is B R2 B' D2 B2 D2 R U' L' B' D' R U F R' F U2 L' U2. [alg.cubing.net]

It took me a few hours to figure all of this out, considering that I had to determine the condition on the set of positions which can and cannot be conjugated to be inverted, but if you would like me to do some more examples, give me any 3x3x3 position which obeys the restriction, and I will simply post a sequence B for each one. Clearly considering permutations alone, this can be done for the nxnxn as well.

#### PM 1729

##### Member
Interesting conjecture. Is any part of it easier to prove?
Somehow I feel that proving that no such conjugate exists for some A might be easier.

Also, a silly question: are rotations taken into account as well? I mean: do there exist some conjugate sequences that are possible only with rotations or can rotations be disregarded?

#### Christopher Mowla

Interesting conjecture. Is any part of it easier to prove?
Somehow I feel that proving that no such conjugate exists for some A might be easier.
Actually, I have a proof already planned, and thus I only need write it out. I wrote "conjecture" simply because I chose not to post a proof at this time.

Also, a silly question: are rotations taken into account as well? I mean: do there exist some conjugate sequences that are possible only with rotations or can rotations be disregarded?
The real hindrance for there not existing such conjugates for inverse is strictly related to corner and middle edge orientations. Since rotations do not affect how many orientations are in each cycle (they just change the orientation of the entire puzzle), they can be disregarded.

#### qqwref

##### Member
cmowla, you mention "positions in which their order is equal to the least common multiple of all their disjoint cycles of corners and middle edges". What would make this be the case as opposed to not be the case? Your scramble A seems to have a 4-cycle of edges, a 7-cycle of edges, a twisted 4-cycle of corners, and a twisted 3-cycle of corners. So the order would be LCM(4, 7, 4*3, 3*3) = 252 (the *3s being for the twisted cycles) but the LCM of the disjoint cycles would be, I assume, LCM(4, 7, 4, 3) = 84. If the edges were instead a 2-cycle and a 9-cycle the order would be LCM(2, 9, 4*3, 3*3) = 36 and the LCM would be LCM(2, 9, 4, 3) = 36, the same. However, I think (but cannot be sure without trying) that this scramble with modified edges would not have a sequence B either.

My counter-conjecture is as follows: for a position A, let T+(n) be the number of clockwise-twisted corner cycles of length n (counting a solved corner as a 1-cycle) and T-(n) be the number of counter-clockwise-twisted corner cycles of length n (again counting a solved corner as a 1-cycle). If T+(n) != T-(n) for any n, then there is no position B for which B A B' = A'. Since corners are the only thing with more than 2 orientations, this should apply to all NxNxN cubes, not just the 3x3x3.

Here's some hand-wavy reasoning for why this should be true. Applying a conjugate can not change cycle count, number of pieces in each cycle, or amount of twist, but can only change which pieces are involved. Thus a conjugate can change any non-twisted cycle, or a twisted edge-cycle, to its inverse, by just switching the order of pieces - the inverse of a flipped edge is the same flipped edge, because an edge has 2 orientations. If we conjugate a twisted corner-cycle, we cannot change its twist; but inverting a corner cycle does change its twist, and specifically, inverting a clockwise-twisted cycle gives you a counter-clockwise-twisted cycle and vice versa. So T+(n) and T-(n) are invariant over conjugation, but inverting A will swap T+(n) and T-(n). Thus if a position A has an n for which T-(n) and T+(n) are not the same, then any position B A B' will have the same T+(n) as A, but A' will have a different T+(n), and thus B A B' and A' can't be the same for any B.

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#### Christopher Mowla

cmowla, you mention "positions in which their order is equal to the least common multiple of all their disjoint cycles of corners and middle edges". What would make this be the case as opposed to not be the case?
It comes down to the fact that you cannot twist a single corner...that's all I can say at this moment in time...but I'm guessing that you already know why based on your "counter-conjecture", which actually agrees with my conjecture (see the end of this post), although I do say that my way to prove this is very different.

Your scramble A seems to have a 4-cycle of edges, a 7-cycle of edges, a twisted 4-cycle of corners, and a twisted 3-cycle of corners. So the order would be LCM(4, 7, 4*3, 3*3) = 252 (the *3s being for the twisted cycles) but the LCM of the disjoint cycles would be, I assume, LCM(4, 7, 4, 3) = 84.
That's correct (partly). In my example I gave two scrambles: the one with a twisted 4-cycle and a twisted 3-cycle of corners and a 7-cycle and 4-cycle of edges. Order = 252. I claimed that this scramble cannot be inverted by conjugation due to the twisted corner cycles. I then modified this scramble to create a scramble which has a 4-cycle and a 3-cycle of corners and a 7-cycle and 4-cycle of edges, and I inverted it with conjugation. Order = 84. Did you miss that?

If the edges were instead a 2-cycle and a 9-cycle the order would be LCM(2, 9, 4*3, 3*3) = 36 and the LCM would be LCM(2, 9, 4, 3) = 36, the same.
I don't debate that...I stated the if the order is larger than the LCM (I was comparing the order to the LCM, not LCM to LCM). In your terms, if there exists any twisted corner cycles or edge cycles (and like you said, counting a solved but unoriented corner or edge as a twisted 1-cycle), then that position cannot be inverted by conjugation. Otherwise, it can be.

However, I think (but cannot be sure without trying) that this scramble with modified edges would not have a sequence B either.
Again, I think you clearly missed my modified algorithm which I found the inverse by conjugation. The edge position is fine. The corner orbit, due to the fact that it had twisted cycles, was a problem.

My counter-conjecture is as follows: for a position A, let T+(n) be the number of clockwise-twisted corner cycles of length n (counting a solved corner as a 1-cycle) and T-(n) be the number of counter-clockwise-twisted corner cycles of length n (again counting a solved corner as a 1-cycle). If T+(n) != T-(n) for any n, then there is no position B for which B A B' = A'.
I don't see how this is a counter-conjecture, as my conjecture claims that only if T+(n) = T-(n) = 0 for corners and T+(n) = 0 for middle edges can a position A have multiple positions B such that B A B' = A'.

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#### Lucas Garron

##### Moderator
Staff member
It comes down to the fact that you cannot twist a single corner...
No.

Suppose we allow any corner twist as a valid move, and consider a position that has a clockwise 8-cycle of corners (doesn't matter what the edges are).
There is no way to turn this into its inverse (which has a counter-clockwise 8-cycle of corners) using a conjugate.

I don't see how this is a counter-conjecture, as my conjecture claims that only if T+(n) = T-(n) = 0 for corners and T+(n) = 0 for middle edges can a position A have multiple positions B such that B A B' = A'.
If there is a conjugate, it is trivial that there are multiple ones. (For example, replace B with B + superflip.)

#### qqwref

##### Member
That's correct (partly). In my example I gave two scrambles: the one with a twisted 4-cycle and a twisted 3-cycle of corners and a 7-cycle and 4-cycle of edges. Order = 252. I claimed that this scramble cannot be inverted by conjugation due to the twisted corner cycles. I then modified this scramble to create a scramble which has a 4-cycle and a 3-cycle of corners and a 7-cycle and 4-cycle of edges, and I inverted it with conjugation. Order = 84. Did you miss that?

I don't debate that...I stated the if the order is larger than the LCM (I was comparing the order to the LCM, not LCM to LCM).
I saw the modified A, but just didn't bother to figure out its cycles I'm not sure what you mean by "comparing the order to the LCM, not LCM to LCM" though, as in my second example (which I don't have a scramble for, would it help if I made one?) the order is 36 and the LCM is also 36, and yet it should not be invertable by conjugation.

Again, I think you clearly missed my modified algorithm which I found the inverse by conjugation. The edge position is fine. The corner orbit, due to the fact that it had twisted cycles, was a problem.
Yep, edges don't matter. My point in the 2-and-9-edge-cycle case is that it's a case which gives different answers for our two conjectures: yours says the order is the same as the LCM, so it should be solvable; mine says T+(4) != T-(4) so it should not be solvable.

I don't see how this is a counter-conjecture, as my conjecture claims that only if T+(n) = T-(n) = 0 for corners and T+(n) = 0 for middle edges can a position A have multiple positions B such that B A B' = A'.
T+(n) = 0 for edges is irrelevant. Also, I don't think having multiple positions B matters, but lgarron showed there are always at least 2 (if there is one at all)

#### Christopher Mowla

No.

Suppose we allow any corner twist as a valid move, and consider a position that has a clockwise 8-cycle of corners (doesn't matter what the edges are).
There is no way to turn this into its inverse (which has a counter-clockwise 8-cycle of corners) using a conjugate.
I'm not exactly sure what a "clockwise 8-cycle of corners" is. Can you either give me an example scramble or a more precise description?

If there is a conjugate, it is trivial that there are multiple ones. (For example, replace B with B + superflip.)
As always, when I mention something like this, I'm referring to non-trivial cases, but that is a nice observation nevertheless.

I saw the modified A, but just didn't bother to figure out its cycles I'm not sure what you mean by "comparing the order to the LCM, not LCM to LCM" though, as in my second example (which I don't have a scramble for, would it help if I made one?) the order is 36 and the LCM is also 36, and yet it should not be invertable by conjugation.
You're right. I didn't read your post correctly. I need to rephrase that part of my conjecture to be what I told you:
...if there exists any twisted corner cycles or edge cycles (and like you said, counting a solved but unoriented corner or edge as a twisted 1-cycle), then that position cannot be inverted by conjugation. Otherwise, it can be.
Yep, edges don't matter. My point in the 2-and-9-edge-cycle case is that it's a case which gives different answers for our two conjectures: yours says the order is the same as the LCM, so it should be solvable; mine says T+(4) != T-(4) so it should not be solvable.
I now see that you're correct. What confused me was that I was thinking about parity conflicts.

Speaking of which, I believe the following condition must exist for permutations (so the above quote of my previous post was the condition for orientations). If someone finds a counterexample, I would appreciate it!

Based on my method (which I cannot modify, as inverse by conjugation is a simple process), if a nxnxn cube position has either a {5,3,3} or {3,3,3,3} of middle edges mixed with either a {7} or {5,3} of corners, then I cannot find a B for these cases, regardless whether or not they have twisted corner cycles.
The permutation part of B for corners can always be an odd permutation, but it must be an odd permutation for the {7} or {5,3}. The permutation part of B for middle edges must always be an even permutation for the {5,3,3} and {3,3,3,3}.

For the nxnxn supercube, the following 19 cycle types must be an even permutation:
{{9,7,7},{9,9,5},{11,7,5},{11,9,3},{13,5,5},{13,7,3},{15,5,3},{17,3,3},{7,7,5,5},{7,7,7,3},{9,5,5,5},{9,7,5,3},{9,9,3,3},{11,5,5,3},{11,7,3,3},{13,5,3,3},{15,3,3,3},{5,3,3,3,3,3,3},{3,3,3,3,3,3,3,3}},

and the permutation portion of B for all of the following 14 cycle types must be an odd permutation:

{{23},{13,11},{15,9},{17,7},{19,5},{21,3},{5,5,5,5,3},{7,5,5,3,3},{7,7,3,3,3},{9,5,3,3,3},{11,3,3,3,3},{5,5,5,3,3,3},{7,5,3,3,3,3},{9,3,3,3,3,3}}.

(Notice that no odd permutation cycle type is listed above because if a cycle type contains even a single even-cycle, then that cycle type can be setup as either an even or odd permutation.)

I believe these cycle types will pose no conflict for the 6 colored nxnxn cube because the parity of wing edges is independent of corners, and we can assume that the parity of any given non-fixed center orbit is either even or odd to our needs, due to same color centers.
Therefore the only permutation conflict I can see at this point in time is {5,3,3} or {3,3,3,3} of middle edges mixed with either a {7} or {5,3} of corners, and if any of these four permutation combinations exist, then middle edge orientations do matter, but since these parity conflicts are excluded from positions such that B A B' = A' for some B, then for all positions A such that B A B' = A', for some B, edge orientations should never matter.

In short, here is my revised conjecture.
[HR][/HR]My Conjecture (Revised)
For all 6 colored even nxnxn cube positions A in which the corner orientation sum is not equal to 0 (mod 3) in all disjoint cycles (where we consider all correctly placed corners to be in a single "cycle"), there are no positions B such that B A B' = A'.

For all 6 colored odd nxnxn cube positions A in which the corner orientation sum is not equal to 0 (mod 3) in all disjoint cycles (where we consider all correctly placed corners to be in a single "cycle") and/or the corner and middle edge permutation combination is either a {7} or {5,3} of corners mixed with either a {5,3,3} or {3,3,3,3} of middle edges, there are no positions B such that B A B' = A'.

For all other nxnxn cube positions (not the supercube) A there exists multiple non-trivial positions B such that B A B' = A'.

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#### cuBerBruce

##### Member
The permutation part of B for middle edges must always be an even permutation for the {5,3,3} and {3,3,3,3}.
This appears not to be true:

Code:
gap> g := (2,3)(4,8)(5,7)(6,9)(10,11);
(2,3)(4,8)(5,7)(6,9)(10,11)
gap> (g^-1)*(1,2,3)(4,5,6)(7,8,9)(10,11,12)*g;
(1,3,2)(4,6,5)(7,9,8)(10,12,11)

gap> g := (1,4)(2,3)(6,10)(7,9)(8,11);
(1,4)(2,3)(6,10)(7,9)(8,11)
gap> (g^-1)*(1,2,3,4,5)(6,7,8)(9,10,11)*g;
(1,5,4,3,2)(6,8,7)(9,11,10)

#### qqwref

##### Member
For all 6 colored even nxnxn cube positions A in which the corner orientation sum is not equal to 0 (mod 3) in all disjoint cycles (where we consider all correctly placed corners to be in a single "cycle"), there are no positions B such that B A B' = A'.

For all 6 colored odd nxnxn cube positions A in which the corner orientation sum is not equal to 0 (mod 3) in all disjoint cycles (where we consider all correctly placed corners to be in a single "cycle") and/or the corner and middle edge permutation combination is either a {7} or {5,3} of corners mixed with either a {5,3,3} or {3,3,3,3} of middle edges, there are no positions B such that B A B' = A'.
I still think you CAN have twisted cycles - you just have to have the same number of clockwise and counterclockwise cycles of each length.

Here's a position with T+(2) = T-(2) = 1:
A = R2 F2 R' B' R F2 R2 F R B R' F'.
Let B = R U2 R2 F D' F D F2 R2 U' B U' B' R'.
Then B A B' A = solved.

#### bobthegiraffemonkey

##### Member
I still think you CAN have twisted cycles - you just have to have the same number of clockwise and counterclockwise cycles of each length.

Here's a position with T+(2) = T-(2) = 1:
A = R2 F2 R' B' R F2 R2 F R B R' F'.
Let B = R U2 R2 F D' F D F2 R2 U' B U' B' R'.
Then B A B' A = solved.
There are easier examples you know
A=[R, U]
B=[U' L2 U L' : F2]

A=[S' R S, U]
B=F' U2 F