# Intro to group theory



## hawkmp4 (Jul 1, 2008)

I have a pretty good background in math (for a high school student) I've taken AP Calc AB.
But all of the sites I've found on group theory are way over my head. Could anyone link me to some kind of resource where I could learn enough group theory to at least get the concepts and apply some to the cube? If not that's fine, it may be just too much at this point in my education.


----------



## cuBerBruce (Jul 1, 2008)

If you haven't tried these sites, you might give them a try. They both talk about group theory in relation to Rubik's Cube and other similar puzzles.

On Jaap's Puzzle Page web site, there is:
http://www.geocities.com/jaapsch/puzzles/groups.htm

It might be a little more advanced, but there is also David Joyner's Rubik's Cube course notes, that now can be found at this location:
http://www.permutationpuzzles.org/rubik/webnotes/

An extended version of this is available as a book (Adventures in Group Theory).


----------



## It3ration (Jul 1, 2008)

I'd recommend the Ryan Heise method. He explains basic group theory and it's applications in cubing on his site.

http://www.ryanheise.com/cube/theory.html


----------



## badmephisto (Jul 1, 2008)

I would recommend you get an actual Book, not an internet site. These kinds of things are better to study that way. Look for a library close to you or something with a math section.


----------



## ubervern (Jul 2, 2008)

I just finished a BS in math with 6 credit hours specifically in group theory, and for almost all of it I have ever seen I either cannot understand it or I do not see how it would help anyone solve a cube. 
Permutations and cycles are applicable but they are not really group theory.
The problem with a cube in group theory is that it only qualifies as a group via permutations, and there is no simple relation between a face turn and the effect it has upon a permutation (computers can track that stuff but not me). The first rule of a group is that it must be associative, ie (a • b) • c = a • (b • c) but this is not true in general for face turns of a cube, thus face turns are not the operators of a rubik's group and group theory doesn't help a solution very much, but since any permutation is a group, there are a few simple observations that can be made, as Ryan states clearly on his site. (actually I didn't see one of the most reassuring statements from group theory: as a finite group a rubik's cube is guaranteed to have a solution in a finite number of steps)

There is highly developed group theory for a cube and it can help discover new algs and optimal solutions, but in general it does not help someone solve a cube.

If you really are interested in group theory, I recommend Abstract Algebra by I.N. Herstein, a very compact yet thorough text.


----------



## AvGalen (Jul 2, 2008)

ubervern said:


> I just finished a BS in math with 6 credit hours specifically in group theory, and for almost all of it I have ever seen I either cannot understand it or I do not see how it would help anyone solve a cube.
> Permutations and cycles are applicable but they are not really group theory.
> The problem with a cube in group theory is that it only qualifies as a group via permutations, and there is no simple relation between a face turn and the effect it has upon a permutation (computers can track that stuff but not me). The first rule of a group is that it must be associative, ie (a • b) • c = a • (b • c) but this is not true in general for face turns of a cube, thus face turns are not the operators of a rubik's group and group theory doesn't help a solution very much, but since any permutation is a group, there are a few simple observations that can be made, as Ryan states clearly on his site. (actually I didn't see one of the most reassuring statements from group theory: as a finite group a rubik's cube is guaranteed to have a solution in a finite number of steps)
> 
> ...


Group theory is indeed not (very) usefull for finding a brute-force 1 step solution to the cube and it won't help a beginner to solve a cube. However, you have stated this yourself


> ...The first rule of a group is that it must be associative


. I don't really understand what that means from a formal math point of view, but for me it means that you can apply group theory if you create independent groups. And that is actually the base of how we solve the cube most of the time. (Cross/F2L/OLL/PLL can be considered independent groups)

Commutators are another obvious example where you divide the cube in several groups. Having a P-part that changes only the 1st group, have a Q-part that only changes the 2nd group and undoing P and Q so the result would still be a solved cube. If P and Q only have a very small intersection (are almost non-associative) the effect on the entire cube will be very small.

And finally group part plays a very big role in really seperate groups that are used for the 2-phase solve of Cube Explorer, The human tistlewait method and most blindfolded methods.

Almost every puzzle is solved by dividing it in groups that have either no influence on each other (think blindfolded), or you just don't care about the influence (think reduction).


----------



## ubervern (Jul 2, 2008)

I certainly agree, but these things you call groups have very little in common with mathematical group theory. Most often people wish to assign a group to a cube state that can be obtained through specific face turns, such as all positions obtainable while restricted to L2 and R2 turns. Sometimes people assign groups via block building methods, but these are not necessarily mathematical groups and learning group theory will not help you understand these things. Cycles and permutations can help, with commutators being a method of solving a permutation, and these things are often discussed in group theory but are not restrained by the rigid definitions of a group.
Basically I am claiming that mathematicians defined the word group in such a way as to not allow people to randomly assign a section of their cube as a group. Does that make sense?


----------



## Lucas Garron (Jul 2, 2008)

ubervern said:


> such as all positions obtainable while restricted to L2 and R2 turns.


<L2, R2>
I love that 2-gen!



ubervern said:


> Basically I am claiming that mathematicians defined the word group in such a way as to not allow people to randomly assign a section of their cube as a group. Does that make sense?


What's that mean. U, F, and R generate a group, right?

And who can stop me from calling the M-slice a group of 8 pieces? 
(_Arbitrarily_, people, not _randomly_!)


----------



## ubervern (Jul 2, 2008)

If you took a solved cube and limited the face turns, such as only doing M turns, you would have a sub-group of the rubiks group, so you can call the M slice a group, but it isn't very interesting. If you took a scrambled cube and limited face turns, then you obtain a subgroup of a permutation of the rubrik's group, but since it does not necessarily contain the solved position, which is the identity of the rubrik's group, it is not necessarily a sub-group of the rubrik's group. I think the parity issues of blind solves and larger cube solves exemplify this perfectly.


----------



## badmephisto (Jul 2, 2008)

well you can use the concepts from group theory to prove all kinds of useful (non-intuitive) conjectures about rubik's cubes, such as the fact that you cannot make odd number of swaps. I wouldn't go as far as to suggest that it is not something worth looking into.


----------



## cuBerBruce (Jul 3, 2008)

ubervern said:


> The problem with a cube in group theory is that it only qualifies as a group via permutations, and there is no simple relation between a face turn and the effect it has upon a permutation (computers can track that stuff but not me).



I'm not sure what you are really getting at here.

From Cayley's theorem (see http://en.wikipedia.org/wiki/Cayley's_theorem ), it is known that all groups can be thought of as equivalent to some permutation group. The Rubik's cube group can be considered equivalent to a permutation group describing how the of the 48 corner and edge "facelets" (mathematicians tend to use the term "facet") of the cube can be permuted.



ubervern said:


> The first rule of a group is that it must be associative, ie (a • b) • c = a • (b • c) but this is not true in general for face turns of a cube, ...


Huh? Successive application of face turns is definitely associative. (U F) R = U (F R), for example. Doing U followed by F R is the same as doing U F followed by R. It is associative! Incidently, sometimes the group operation for Rubik's cube has been described as "followed by." U F R means U followed by F followed by R. (This seems to be rather informal to me. The group operation basically describes how an ordered pair of elements of the group can be combined to produce another element of the group.)


----------



## ubervern (Jul 3, 2008)

order of operations, PEMDAS: (U F) R means do U then F, then R. U (F R) means do parenthesis first, so do F then R then do U, clearly non-associative. If you used the "is followed by" method then associativity would be trivial, everything would be associative. In most systems (such as typical multiplication) you can cheat order of operations and continue working left to right, but this is because the operation is commutative, which is again not true for the rubrik's cube (F R is not the same as R F). When you look at U (F R) and do U first you have just cheated order of operations and commutated the elements. It will never get you in trouble with real numbers, but is quite wrong here.

All groups are isomorphic to some permutation group, but the "rubrik's group" is a permutation group, so again the statement is trivial. The only group operation I know of for the cube is a permutation. These confusions are the basis of my point, group theory is not easily nor naturally applied to a cube.


----------



## badmephisto (Jul 3, 2008)

ubervern said:


> order of operations, PEMDAS: (U F) R means do U then F, then R. U (F R) means do parenthesis first, so do F then R then do U, clearly non-associative. If you used the "is followed by" method then associativity would be trivial, everything would be associative. In most systems (such as typical multiplication) you can cheat order of operations and continue working left to right, but this is because the operation is commutative, which is again not true for the rubrik's cube (F R is not the same as R F). When you look at U (F R) and do U first you have just cheated order of operations and commutated the elements. It will never get you in trouble with real numbers, but is quite wrong here.
> 
> All groups are isomorphic to some permutation group, but the "rubrik's group" is a permutation group, so again the statement is trivial. The only group operation I know of for the cube is a permutation. These confusions are the basis of my point, group theory is not easily nor naturally applied to a cube.



I could be wrong but I am going to side with cuberbruce on this one. It seems to me like you are confusing commutativity with associativity. And it is not true that associativity is trivial in all circumstances. Consider this simple example:
5-(3-2) != (5-3)-2

Something that means do F R first then U is simply just F R U, not U (F R). In other words it is my belief that moves on the rubik's cube are indeed associative but need not commute.


----------



## JBCM627 (Jul 4, 2008)

badmephisto said:


> ubervern said:
> 
> 
> > order of operations, PEMDAS: (U F) R means do U then F, then R. U (F R) means do parenthesis first, so do F then R then do U, clearly non-associative.
> ...



Ubervern or someone correct me if I am wrong. I think his point was that you still need to stick to order of operations when considering associativity. Operations in parentheses are always done first in a math equation; you simply have the luxury of ignoring this when working with pure numbers.

This may seem counterintuitive, as algorithms are time dependant, and so you are also used to reading cubing algorithms from left to right and generally ignoring ()'s. I think ubervern is considering permutations instead of algorithms because of this. As permutations are not time dependant, think of this as the end result of executing the moves A(BC), referring to order of operations as a reference for what moves to perform first; 'left to right' taking lower precedence than the order in which you should do the operations.

Of course, this is from a purely mathematical standpoint. You would *not* do this while reading it as an algorithm (time dependent), instead of a permutation (time independent).

So for practical purposes, we can define our own operations, based on time, giving it precedence over all other operations. Now operations should be read in chronological order, before considering temporally invariant operations. If the operator "A•B" means "Do A before considering B", then A•B•C translates to: do A, then do B, then do C. A•(B•C) means the same thing, as does (A•B)•C. Thus algorithms are associative.


----------



## Swordsman Kirby (Jul 4, 2008)

ubervern said:


> "rubrik's group"



Stop it!!!


----------



## JBCM627 (Jul 4, 2008)

Swordsman Kirby said:


> ubervern said:
> 
> 
> > "rubrik's group"
> ...



Yes indeed.


----------



## Stefan (Jul 4, 2008)

ubervern said:


> U (F R) means do parenthesis first, so do F then R then do U, clearly non-associative.


This caused my best laugh today. I have this theory that your "BS" in math doesn't stand for bachelor of science.


----------



## ubervern (Jul 7, 2008)

StefanPochmann said:


> ubervern said:
> 
> 
> > U (F R) means do parenthesis first, so do F then R then do U, clearly non-associative.
> ...



Perhaps I am not very good at making clear explanations, but that is no reason to throw out insults. 

Let me explain this in as simple terms as I can:
First a set is defined. This can be any list of objects, preferably with something in common. 
Then an operation is defined. This can be anything from simple addition to the assembly of a computer. 
Once these are established it is possible to begin to determine if the set with the defined operation satisfies the requirements of a group. 

Associativity is defined, as I stated earlier, (X*Y)*Z = X*(Y*Z). A set, with the defined operation, is considered associative if it satisfies the above for ALL X, Y, Z in the set. 
Example: addition is associative, (5+2)+3 = 5+(2+3) True
Example: division is NON-associative, (6/3)/2 = 6/(3/2) false. For the Left hand side, 6/3 is 2. 2/2 is one. For the right, 6 is to be divided by 3/2, or 1.5, yielding 4.

For the cube, it becomes difficult to define the specific operation using face turns, simply because group theory does not formally allow different operations such as F, R, and U, or even +, -, x. (this is yet another reason it is difficult to apply group theory to a cube). I am uncertain how to get around this, but assuming it was figured out, the face turns are still non associative.


----------



## JBCM627 (Jul 7, 2008)

Ok, children... be nice. Uber, read this.


----------



## cuBerBruce (Jul 8, 2008)

ubervern said:


> For the cube, it becomes difficult to define the specific operation using face turns, simply because group theory does not formally allow different operations such as F, R, and U, or even +, -, x. (this is yet another reason it is difficult to apply group theory to a cube). I am uncertain how to get around this, but assuming it was figured out, the face turns are still non associative.



ubervern, since you see U, F, R, etc. as "operations," you seem to be trying to relate them to an operation for a group. This seems to be why you don't see the natural relationship between Rubik's Cube and group theory.

Two things that are necessary to have a group are a set and an operation (the "group operation"). And yes, you are correct in that a group has *one* group operation, not several. In the Rubik's Cube group, the "operations" U, F, R, D, B, and L are not group operations, but elements of the set for the group. (There are 43252003274489855994 other "operations" in this set, but I won't bother listing them out.  ) These operations can be thought of as "permutations." (Each such set element basically describes a way the "facelets" of Rubik's Cube can be permuted.)

Permutations can be composed. That is, a permutation can be applied to the result of another permutation. The net result of composing the two permutations (and the order you compose them is important, of course) is a third permutation. The group operation for the Rubik's Cube group is basically what I just described: composition of permutations. In this post, I will explicitly use an asterisk (*) to represent the operator for the group operation, just to be clear that the group operation is a separate concept from the face turns.

So F*R represents the result of composing R with F. U*F represents the result of composing F with U. The expression (U*F)*R means the result of composing R with the result of composing F with U. The expression U*(F*R) means the result of composing R with F, and then composing the resulting permutation with U. It happens that composition of permutations is associative. Both U*(F*R) and (U*F)*R result in the same permutation. So basically we don't need to bother with the parentheses and simply write U*F*R.


----------



## JBCM627 (Jul 8, 2008)

cuBerBruce said:


> ubervern, since you see U, F, R, etc. as "operations," you seem to be trying to relate them to an operation for a group. This seems to be why you don't see the natural relationship between Rubik's Cube and group theory.



Yet, you can still consider U, F, and R (etc) as operations, correct? It is another way of thinking about it, perhaps, but still legitimate. It even makes more intuitive sense to consider it this way, as turns seem to be operations you do to a cube.

More concretely, consider the cube a tensor with each element representative of a single cubie on the cube. All the cubies may be assigned a unique value that represents position and orientation. Then you may define U, F, R, etc as transformations which can be applied to your tensor through simple multiplication.

This will still work out to be associative, as tensor multiplication is. If you were to apply the algorithm U R F on state S, you would have: F*R*U*S. You can do (F*R)*U*S, and it will be the same as F*(R*U)*S.

Edit: Interestingly, as you only need to turn 5 faces to solve a cube, you will only need to use 5 operations. In (multi)linear algebra terms, you can show that the system of operations is dependent, and can thus determine that U is a combination of R, F, D, B, and L quite easily.

Edit again: Although, I just realized I'm not sure whether the transformations would be considered operations; could it be only the multiplication? Or the two combined? I think the two combined, as the transformations require both the transformation tensor and the multiplication definition to work.


----------



## ubervern (Jul 8, 2008)

yall explain things much better than me.

If we number the top four corners going clockwise, then a U turn takes 1 and sends it to 2, 2 to 3, 3 to 4, and 4 back to 1.
A permutation of this would be 2341. But the permutation is specific, it always sends 1 to 2 and so on, while a U turn is arbitrary, what it really does is send whatever happens to be at 1 to 2, whatever happens to be at 2 to 3, and so on. 
It has been my belief that these are not always the same thing but I have yet to find an example where they differ. If these are the always same then I concede.


----------



## Swordsman Kirby (Jul 8, 2008)

ubervern said:


> But the permutation is specific, it always sends 1 to 2 and so on,



Well, that's wrong. Your explanation for an "arbitrary" U turn is actually describing a permutation.


----------



## JBCM627 (Jul 8, 2008)

ubervern said:


> yall explain things much better than me.
> 
> If we number the top four corners going clockwise, then a U turn takes 1 and sends it to 2, 2 to 3, 3 to 4, and 4 back to 1.
> A permutation of this would be 2341. But the permutation is specific, it always sends 1 to 2 and so on, while a U turn is arbitrary, what it really does is send whatever happens to be at 1 to 2, whatever happens to be at 2 to 3, and so on.
> It has been my belief that these are not always the same thing but I have yet to find an example where they differ. If these are the always same then I concede.



1->2, 2->3, 3->4, 4->1 is just one specific case of turning the arbitrary corners at your positions 1..4. If I am filling in the blanks properly, it is just the specific case where you are applying a U turn to a cube with those corners solved. They aren't the "same" per se, as there is not a biimplication between the two cases: the specific permutation does not imply existence of a general transformation for arbitrary corners, even though the converse may be true.

Edit: Applying my earlier reasoning, let U be a transformation matrix such that:
U =
0 0 0 1
1 0 0 0
0 1 0 0
0 0 1 0
and let S be your state where:
S = 
a
b
c
d

Now, when a=1, b=2, c=3, and d=4, (where 1-4 are the numbers assigned to specific corners) you have the specific case you mentioned. However a..d can be any corners you wish. U*S will be your permutation. If S' = U*S, then S' =
d
a
b
c

So, the transformation implies the specific case is true. However, let:
S =
1
2
3
4
And we are not able to determine a reasonable U to change it to 2341 without being given more information. For all we know, U could be:
-6 0 0 2
0 .5 0 .5
0 2 0 0
0 0 0 .25
Thus, the reverse implication will not be true unless we are given more information: if we know that each corner must go to one and only one other, we can then derrive the inital U, and would have an equality between the two statements.

Edit again:
For this to work with successive applications of transformations, we do also need to note the order in which we perform these transformations. As in, what I was saying earlier about the need for these operations to be well-ordered (temporally, in our case) needs to hold. As noted, these operations are not commutative.


----------



## Stefan (Jul 8, 2008)

ubervern said:


> StefanPochmann said:
> 
> 
> > ubervern said:
> ...


You're not explaining something badly, your stating something completely wrong. That in combination with the _"BS in math with 6 credit hours specifically in group theory"_ introduction is plenty reason for mockery. To me, at least.

I think it gets more clear if you stop being lazy and write the operator explicity. So not "U (F R)" but "U*(F*R)". The parentheses do *not* mean "do F first, then...". They mean "combine F and R first" (using the group operation).

Also, why did you say "do F then R"? Why not R first? Because you're doing from left to right, aren't you? But then why would you do (F R) before U, when U stands on the left and should be done first? This is so obviously flawed you don't even need a math background or know the cube group to see it.


----------



## F.P. (Jul 8, 2008)

@threadstarter:

I recommend you to read books about it instead of searching for help on the internet.
Most people only have superficial knowledge - and aren't aware of it.

So if you want to learn it right - don't use the internet.


----------



## JBCM627 (Jul 8, 2008)

F.P. said:


> @threadstarter:
> 
> I recommend you to read books about it instead of searching for help on the internet.
> Most people only have superficial knowledge - and aren't aware of it.
> ...



Books may have the advantage of being thorough, but there are some sites that do a really good job explaining, often better than a textbook would as you can pool resources. Wikipedia _is_ an amazing math resource... and Jaaps group theory for puzzles page is a good way to get an introduction to group theory and how it applies to puzzles imho. Check out the bibliography for books on group theory as it pertains to the cube, too.


----------



## F.P. (Jul 8, 2008)

Some certain "serious" sites might be good but especially when asking such questions on a public forum, you can't expect to get useful answers.

Other than that...I personally prefer to learn/read from books and not my pc.


----------



## AvGalen (Jul 9, 2008)

F.P. said:


> Some certain "serious" sites might be good but especially when asking such questions on a public forum, you can't expect to get useful answers.
> 
> Other than that...I personally prefer to learn/read from books and not my pc.


 
Isn't this a public forum :confused:
I expect a useful answer


----------



## F.P. (Jul 10, 2008)

Yep...it's a public forum...erm - that's what I said.

And of course there is the possibility to get useful answers, but I wouldn't count on it. 
At least you should use other sources as well...


----------



## hawkmp4 (Jul 10, 2008)

Thank you. Any books you can recommend? or should I just peruse the math section at the library (I don't mind, I'll just end up getting 3x as many books as I originally wanted xD)

And if I'm following this thread correctly...
U*F*R would apply the transformations U, F, and R consecutively to the cube.
U*(F*R) would apply the transformation U, then the composition of F and R, to the cube... which is exactly the same as U*F*R, at least in the example of a cube?


----------



## Stefan (Jul 10, 2008)

Not quite. U*F*R is the composition of U, F and R. It doesn't apply anything to the cube, but it can be applied to a cube(*). Subtle difference. But yes, applying it to a cube is the same as applying U, F and R to the cube in that order(*). Same with U*(F*R) which is exactly the same as U*F*R, except the latter is "sloppy" notation which is only ok because the composition is associative.

(*) Although "applying it to a cube" is somewhat dirty talk, because what's "a cube"? Is it a "state" of the cube, and is that something different than an "action" that can be applied to the cube? No need to talk about two different things, it's sufficient to talk in terms of cube group elements, which can be interpreted as both actions and states. Buy one, get two.

(**) Although this could be very well be defined the other way around, too. Usually mathematicians would think of U*F*R as the composition of functions U, F and R, and applying it to some input x would usually mean (U*F*R)(x) = U(F(R(x))), so that R gets applied first. Both directions of writing/reading are equally valid, and some groups of people prefer one while others prefer the other. Depends on what suits their needs better. Speedcubers tend to not understand or care about the group theory that can be seen in the cube, and simply write "U F R" meaning instructions supposed to be read and applied left to right.

Then again, I'm only a kind of hobby mathematician. Maybe some of the math gurus can confirm or correct my ramblings?


----------



## JBCM627 (Jul 10, 2008)

StefanPochmann said:


> (*) Although "applying it to a cube" is somewhat dirty talk, because what's "a cube"? Is it a "state" of the cube, and is that something different than an "action" that can be applied to the cube? No need to talk about two different things, it's sufficient to talk in terms of cube group elements, which can be interpreted as both actions and states. Buy one, get two.


I wouldn't consider myself a math guru, but this sounds right to me. You can consider U F R either a transformation on a (solved) cube state x, or simply an element of a group attainable by doing U F R.



StefanPochmann said:


> (**) Although this could be very well be defined the other way around, too. Usually mathematicians would think of U*F*R as the composition of functions U, F and R, and applying it to some input x would usually mean (U*F*R)(x) = U(F(R(x))), so that R gets applied first. Both directions of writing/reading are equally valid, and some groups of people prefer one while others prefer the other. Depends on what suits their needs better. Speedcubers tend to not understand or care about the group theory that can be seen in the cube, and simply write "U F R" meaning instructions supposed to be read and applied left to right.


Slight nuance here that I think should be noted... with the algorithm U F R, applying U then F then R would require you to do U(x) first, then F(U(x)) then R(F(U(x)). So inside out. So if * is an operator denoting order of transformations (instead of composition of functions), (U*F*R)(x) = R(F(U(x))). If * is composition of functions, then what Stefan said is correct. 

Re the earlier associative arguments, note that this is still associative, as (with the composition operator *), U((F*R)(x)) = (U*F)(R(x)).


----------



## hawkmp4 (Jul 11, 2008)

so U is a transformation to the elements of the cube group? I'm just trying to get the terminology and basic concepts down. Sorry if they're stupid questions.


----------



## JBCM627 (Jul 11, 2008)

Consider the elements of the cube group to be all possible permutations (states) of the cube.

You can now "name" these elements. One way of doing so is by stating an algorithm you can apply to a specific permutation, a point of reference, that will get you to the permutation you want to name. So for example consider a solved cube one permutation, one element. I can now say that U will be another permutation, namely a solved cube after I have applied the U algorithm to it. And of course, elements can have more than one name.

In addition to considering U a permutation in the group, you can also consider U a transformation which you may apply to a permutation. Specifically, U is an algorithm that will rearrange the position and orientation of individual cubies on the upper face.

So, more or less, you have (building up):
-the cube, which is just a set containing information in a given arrangement
-transformations, which are the moves themselves, and will rearrange elements of the cube from one position to another
-algorithms, which consist of transformations arranged in a specific order
-permutations, which are positions, or states of the cube, achievable by applying algorithms to a given cube
-the cube group, which is a group with its elements being all permutations of the cube

This is just my personal way of thinking about it based on my experience... there are probably many valid nomenclatures.


----------



## Stefan (Jul 11, 2008)

JBCM627 said:


> StefanPochmann said:
> 
> 
> > (**) Although this could be very well be defined the other way around, too. Usually mathematicians would think of U*F*R as the composition of functions U, F and R, and applying it to some input x would usually mean (U*F*R)(x) = U(F(R(x))), so that R gets applied first. Both directions of writing/reading are equally valid, and some groups of people prefer one while others prefer the other. Depends on what suits their needs better. Speedcubers tend to not understand or care about the group theory that can be seen in the cube, and simply write "U F R" meaning instructions supposed to be read and applied left to right.
> ...


That's kind of what I meant with both directions being possible and valid. Although, I disagree that function composition can't get you (U*F*R)(x) = R(F(U(x))). Simply define the composition A*B of two functions A and B as the function with (A*B)(x) = B(A(x)).

Associativity proof:
((U*F)*R)(x) = R((U*F)(x)) = R(F(U(x))) = (F*R)(U(x)) = (U*(F*R))(x)

And now that writing U*F*R is legitimate, the above line also shows:
(U*F*R)(x) = R(F(U(x)))


----------



## mrCage (Jul 16, 2008)

Hi 

Back in 1984 i could easily understand many concepts of group theory solely from my cube experiences. That said, there are also many concepts of group theory not directly applicable to the cube. Ths lack of applicability of mathematical theory is why i got fed up with pure maths and drifted towards philosophy for a while, and then in the direction of computers and programming. From one extreme to the other i guess. But i like better this latter extreme 

- Per


----------



## qqwref (Nov 5, 2008)

The Rubik's Cube is IMO not the best group to consider if you want to teach someone about group theory. In fact it's one of the worst: to even really understand how big the group is or how to do the group operation on elements of the group, you have to understand quite a lot about how the cube works: that centers are fixed relative to each other; that stickers can't just be moved arbitrarily but are confined to their pieces; which positions are solvable; and how to 'apply' a position to another (which requires a lot of familiarity with permutations, unless you talk about move sequences, which causes a lot of other problems such as understanding equivalent move sequences, creating inverses, numbering the group elements, etc.). It's also not very intuitive because you don't normally think of taking a cube and applying another cube position to it, but rather of taking a cube and applying turns to it, so I'd guess that a lot of people start thinking that the turns are somehow group operations. Since the group is so big it's really hard to get a feel for it, and there's no way you could visualize the entire group. It takes top blindfold cubers ten seconds to even memorize one element of it 

If you want to learn some basic group theory, it would be best to start with simpler examples of groups. The most familiar are probably the groups of addition with integers or with modular arithmetic. Then you can teach permutations (the really useful stuff!) by going into the basics of symmetric groups and cycle notation. As long as you stay away from the Rubik's Cube group, many of the concepts of group theory can be expressed through very simple examples.


----------

