# Intro to group theory

#### JBCM627

##### Member
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.

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#### ubervern

##### Member
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

##### Member
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

##### Member
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.

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#### Stefan

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

##### Member

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

##### Member

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.
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.

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#### F.P.

##### Member
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

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:

#### F.P.

##### Member
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

##### Member
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?

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#### Stefan

##### Member
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?

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#### JBCM627

##### Member
(*) 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.

(**) 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

##### Member
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

##### Member
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

##### Member
(**) 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.
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

##### Member
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

##### Member
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.