Things that are 'n-gen'?

[blah]

"The fundamental theorem of cyclic groups states that if G is a cyclic group of order n then every subgroup of G is cyclic."

EPLL (all 4!/2 cases) is a subgroup of G, and it is easy to see it is not cyclic (every element has order <12). QED.

o i c wat u did thar

elegant proof is elegant

are they really calling it the fundamental theorem of cyclic groups these days?

 

[tx789]

For last layer all you need is <R,U,M> <R,U> for corners and <M,U> for edges. So it's sort of 2 gen

 

[RyanReese09]

For last layer all you need is <R,U,M> <R,U> for corners and <M,U> for edges. So it's sort of 2 gen

Do Aperm on a solved cube.

Using <R, U> How can you solve this? You can't. Your example only works if CP is correct.

 

[Jorghi]

Do Aperm on a solved cube.

Using <R, U> How can you solve this? You can't. Your example only works if CP is correct.

I think you are thinking about 2gll.

 

[macky]

Keep this in mind: in a mathematical context, generators are in terms of positions (piece permutations). A generator like <RUR', D> is valid and useful in a mathematical context because RUR' actually represents a single piece permutation. That generator contains every permutation that can be generated by those two permutations. In the cubing context, though, it's very imporatnt to think in terms of moves themselves, rather than cube positions, because otherwise we can't distinguish betwen R2URUR'U'R'U'R'UR' and F2UR'LF2RL'UF2. So in that spirit we should be thinking of generators in terms of individual moves. Allowing generators of multiple moves just dilutes the definition to the point where there is no meaningful distinction between 2-gen and not. From both a conceptual and a practical viewpoint, it makes more sense to be strict and allow N-gen to only contain individual moves.

tl;dr: We shouldn't consider things like <RUR', D> to be 2-gen.

Yeah, this needs to be understood before even talking about what Bruce wrote.

"The fundamental theorem of cyclic groups states that if G is a cyclic group of order n then every subgroup of G is cyclic."

The refinement is that there's a unique subgroup of order d for every divisor d of n, and that these are all the subgroups of G. So here's an even shorter proof that the cube group is not cyclic: <R> and <U> are distinct subgroups of the same order.

are they really calling it the fundamental theorem of cyclic groups these days?

I wouldn't call it anything since it's "common knowledge."

 

[Meep]

Is there another term that can be used to describe a sequence of turns that involve N faces?

e.g.
Person A: "R U R' D R U' R' D' is 3-gen because it uses 3 different faces."
Person B: "No it's not 3-gen because it can be notated as <R U R', D>, you mean it's 3-______."

I think it'd be easier to just say what faces the alg uses, like "R U R' D R U' R' D' is an RUD alg." It may seem silly, but I personally think it's more easily and readily understood than a more technical term.

 

[Hershey]

I think it'd be easier to just say what faces the alg uses, like "R U R' D R U' R' D' is an RUD alg." It may seem silly, but I personally think it's more easily and readily understood than a more technical term.

"R U R' D R U' R' D is an 3 gen <R,U,D> algorithm"?

 

[Meep]

"R U R' D R U' R' D is an 3 gen <R,U,D> algorithm"?

Yeah, but people were going on about how it could be a 2 gen <R U R',D> algorithm too.

 

[Hershey]

Oh wow, I just noticed that ccw U perm is 4 gen.

R2 U' R' U' R U R U R U' R=
R' R' U' R' U' R U R U R U' R

which uses <R, R' U', R U, R U'>

 

[Sa967St]

Oh wow, I just noticed that ccw U perm is 4 gen.

R2 U' R' U' R U R U R U' R=
R' R' U' R' U' R U R U R U' R

which uses <R, R' U', R U, R U'>

You're doing it wrong.

 

[Hershey]

Oh... Nevermind then.

 

[rcbeyer]

Oh wow, I just noticed that ccw U perm is 4 gen.

R2 U' R' U' R U R U R U' R=
R' R' U' R' U' R U R U R U' R

which uses <R, R' U', R U, R U'>

2-gen can, being as correct as I know, refer also to algs solved/scrambled with <R U R', D> meaning you're allowed to apply the short sequence "R U R'", its inverse, D, or its inverse.

R' R' U' R' U' R U R U R U' R

this alg only consists of R, U and the inverse of them, so this would still be a 2-gen alg of the group <R, U> i believe

 

[uberCuber]

I think you are thinking about 2gll.

2GLL is when corner permutation is already solved. You cannot do an A perm with <R,U,M> (because M moves don't affect the corners, and you can't do an A perm with <R,U>, as evidenced by the fact that there are no A perms in 2GLL)

 

[Erdos]

R' R' U' R' U' R U R U R U' R

this alg only consists of R, U and the inverse of them, so this would still be a 2-gen alg of the group <R, U> i believe

Mathematically, you can still write it as <R' R' U' R' U' R U R U R U' R>, which is a subgroup most likely of order 2. That is, the algorithm is the minimal non-identity element in this generated subgroup. However, you can also write it as <R,U>, which makes it no longer a minimal element in the <R,U> subgroup. In other words, the common way we write it (<R,U>) is not the only way of putting the element (algorithm) in a subgroup (n-gen).

 

[cuBerBruce]

Mathematically, you can still write it as <R' R' U' R' U' R U R U R U' R>, which is a subgroup most likely of order 2. That is, the algorithm is the minimal non-identity element in this generated subgroup. However, you can also write it as <R,U>, which makes it no longer a minimal element in the <R,U> subgroup. In other words, the common way we write it (<R,U>) is not the only way of putting the element (algorithm) in a subgroup (n-gen).

Since <R' R' U' R' U' R U R U R U' R> has a single generator, it must be a cyclic group. The generator has order 3 so the cyclic group it generates also has order 3, meaning the group contains three elements: the generator, the generator squared (which is also the inverse of the generator), and the identity element.

 

[blah]

I wouldn't call it anything since it's "common knowledge."

... since it's *almost trivial. (Most fundamental theorems are "common knowledge," which is why they're called fundamental theorems; however, most of them are non-trivial, which is why they get to be called fundamental theorems. Heh.)

Oh and sweet proof. Exactly what I was looking for.

 

[macky]

I wouldn't call it anything since it's "common knowledge."

... since it's *almost trivial. (Most fundamental theorems are "common knowledge," which is why they're called fundamental theorems; however, most of them are non-trivial, which is why they get to be called fundamental theorems. Heh.)

Yeah, that's more accurate.

 

[Tim Reynolds]

My preferred proof that the cube group isn't cyclic: It's non-abelian. \( a^xa^y=a^ya^x \) but RU is different from UR.

My opinion: If you're looking for a practical definition, there's hardly any reasons to talk about any generators besides the standard ones. People talk about 2-gen algorithms because there's some practical advantage to using <R,U> or <M,U> algorithms. There's no practical reason to talk about <UF,RBL> algorithms. That's why there's a term for <R,U> algorithms: they are a large set of easy-to-execute algorithms with a well-known set of properties. <M,U> falls in the same category. I can't think of any generators that are more than one turn in sets like those.

If you're trying to be cube theoretic, take a step back and try to figure out what you're actually trying to prove. Otherwise you end up with all algs being 1-gen, or something else equally useless.

 

[Erdos]

Since <R' R' U' R' U' R U R U R U' R> has a single generator, it must be a cyclic group. The generator has order 3 so the cyclic group it generates also has order 3, meaning the group contains three elements: the generator, the generator squared (which is also the inverse of the generator), and the identity element.

I actually didn't pay attention to what the actual algorithm is. However, it's quite obvious that if the generator has order 3, the cyclic group (ie. the subgroup created by all elements of the generator) has the same order. That could practically be the definition.

EDIT:

My preferred proof that the cube group isn't cyclic: It's non-abelian. \( a^xa^y=a^ya^x \) but RU is different from UR.

I like this one better too. Proving a group is non-abelian is so much simpler (and trivial) than proving a subgroup isn't cyclic.

 

[blah]

tim reyholds win