What are the group axioms?

[Johnny5]

What are the axioms of a group?

 

I suspect I know them, but i've read conflicting answers from time to time.

[Tom Mattson]

Every presentation of group theory I have seen defines a group <G,*> as a set G with some binary operation * defined on it. And every presentation contains the following 3 group axioms:

 

1. Associativity must hold.

2. There must exist an identity element.

3. There must exist an inverse under * for each element in G.

 

The only possible source of conflict I can imagine is that some references list a fourth axiom:

 

4. G must be closed under *,

 

whereas other references include that as part of the definition of a binary operation on a set, rather than as part of the definition of a group itself.

[Johnny5]

Every presentation of group theory I have seen defines a group <G' date='*> as a set G with some binary operation * defined on it. And every presentation contains the following 3 group axioms:

 

1. Associativity must hold.

2. There must exist an identity element.

3. There must exist an inverse under * for each element in G.

 

The only possible source of conflict I can imagine is that some references list a fourth axiom:

 

4. G must be closed under *,

 

whereas other references include that as part of the definition of a binary operation on a set, rather than as part of the definition of a group itself.[/quote']

 

Ok, thank you. I didn't know that about closure.

 

Is that true for any arbitrary binary relation on a set?

 

 

Let A denote a set, and let R denote a binary relation on A.

 

if x,y elements of A, then x-R-y is an element of A, for any x,y.

 

?

 

Actually, no that cannot be true. Strike the question above.

 

A mathematical operation like multiplication or addition, is not a binary relation. They are "binary operations" as you say which are different.

 

I think then, that it is best to include closure as a group axiom.

[matt grime]

Tom said binary *operation* not binary *relation*, so what was the point of that I mean in what sense can a binary relation even be thought of as producing another element of the set?

 

Anyway, it depends on how they define binary operation. Some leave the codomain implicit some explicit. Often they are are defined to satisfy closure since * will be referred to as a binary operation from GxG to G.

 

Though quite why some 'fault' of binary relations means we ought to state something about a completely unrelated thing is a mystery. (nb no one should ever define functions as relations)

[Johnny5]

Tom said binary *operation* not binary *relation*, so what was the point of that I mean in what sense can a binary relation even be thought of as producing another element of the set?

 

I was thinking of something else, i said strike it. I had something else in mind.

 

Here is what I was thinking of anyways:

 

Suppose you define the binary relation -B- on the set of moments in time T.

 

suppose that x,y are elements of T.

 

Then if the statement

 

x-B-y is true, then since B is a relation on T, it must follow that x,y are elements of T. So that by assigning the binary relation x-B-y the value true, it must be the case that x,y are elements of the set the binary relation was defined on. I just got slightly confused, but this is what i was thinking of.

 

To say what i was thinking of one more time, if you know that R is a relation on set A, and in your work you are busy treating f-R-g as true, it is inferrable just by looking at the expression f-R-g, that f,g are elements of the set A, which R was defined on.

 

Like i said, my mistake. I was thinking of something else.

 

Anyway' date=' it depends on how they define binary operation. Some leave the codomain implicit some explicit. Often they are are defined to satisfy closure since * will be referred to as a binary operation from GxG to G.

 

Though quite why some 'fault' of binary relations means we ought to state something about a completely unrelated thing is a mystery. (nb no one should ever define functions as relations)[/quote']

[matt grime]

suppose that x' date='y are elements of T.

 

Then if the statement

 

x-B-y is true, then since B is a relation on T, it must follow that x,y are elements of T.

 

 

Doesn't that strike you as being completely tautologous? It ought to, since it is. If x and y are not in T then xBy makes absolutely no sense. The domain is part of the definition of relation (and function).

 

Moreover if x and y are not both in T then xBy is neither true nor false - it is out of the domain of B and is meaningless.

[Johnny5]

If x and y are not in T then xBy makes absolutely no sense.

 

Part of the definition, if you make it so. But I understand what you said perfectly, if x,y are not in T, then xBy makes absolutely no sense.

 

Exactly.

 

Treating 'before' in the customary way, to say house before tree is totally meaningless, its not even false, it is meaningless.

 

Just a logical point, which you appear to understand too.

 

Regards Matt

 

Ahh, i read a little further, and I quote you:

 

Moreover if x and y are not both in T then xBy is neither true nor false - it is out of the domain of B and is meaningless.

 

 

I will brush up on the definition of relation.

 

Let A denote a set.

 

Then we can talk about the set A X A, where X is the cartesian product.

 

Now, we can define a binary relation on A, using the cartesian product somehow, i just forget the formalism.

 

Definition- Binary relation

 

Definition (binary relation):

A binary relation from a set A to a set B is a set of ordered pairs <a, b> where a is an element of A and b is an element of B.

When an ordered pair <a, b> is in a relation R, we write a R b, or <a, b> R. It means that element a is related to element b in relation R.

When A = B, we call a relation from A to B a (binary) relation on A .

 

Ok, so a binary relation on A, is a subset of A X A, where X denotes the binary relation "cartesian product" defined on the set of sets.

 

Odd, but there you go.

[matt grime]

Take it to its (il)logical conclusion: if < denotes less than on the real numbers is duck<lion true? No, but neither is its negation: it is meaningless, since < isn't a relation on the animal kingdom.

[Johnny5]

Take it to its (il)logical conclusion: if < denotes less than on the real numbers is duck<lion true? No, but neither is its negation: it is meaningless, since < isn't a relation on the animal kingdom.

 

Exactly. And thinking in this fashion, is highly advanced to say the least.

 

Kind regards Matt

 

PS: Not too many people realize this nuance here, or if they do they certainly don't appreciate its importance.

[Lyssia]

One of my algebra texts defines a binary relation like this, as a subset of the Cartesian product. It seems perfectly logical to me now but when I first read it I was like "Woooooaaaaaaah!"

[Johnny5]

One of my algebra texts defines a binary relation like this, as a subset of the Cartesian product. It seems perfectly logical to me now but when I first read it I was like "Woooooaaaaaaah!"

 

A minor technical point.

 

I think defining a binary relation as a subset of the Cartesian product, leads to the Russell paradox of set theory, or possibly to the set of all sets, which has the Russell problem.

 

I haven't worked it out yet, I'm just gonna save it for later.

 

Regards

[matt grime]

How? I'd put money on it not doing.

[Johnny5]

How? I'd put money on it not doing.

 

Well i saw the potential for the problem there yesterday.

[matt grime]

So what? No disrespect, but I do not trust your judgement on matters mathematical. A binary relation requires sets A, B, AxB, and a subset of AxB. I see nothing about a set of all sets being mentioned (which is Cantor's Paradox - Russell is the set of sets that do not contain themselves)

[Johnny5]

So what? No disrespect, but I do not trust your judgement on matters mathematical. A binary relation requires sets A, B, AxB, and a subset of AxB. I see nothing about a set of all sets being mentioned (which is Cantor's Paradox - Russell is the set of sets that do not contain themselves)

 

And you should not trust me on matters mathematical, thats what binary logic is for.

 

so anyhow...

 

ok set of all sets Cantor's paradox

 

Also, Russell's paradox is the set of sets which are not elements of themselves. I knew he found that one.

 

But the two are related, i discovered that some time ago.

 

Moving on

 

Let me have a look at what you say here...

 

 

A binary relation requires sets A, B, AxB, and a subset of AxB.

 

Let me see if i can develop the problem...

 

Definition:

For any set A, and any set B: AXB ={(x,y)| x in A AND y in B}

 

The LHS is called the cartesian product of set A with set B.

 

e.g.

 

Permit A={1,2,3,4}

Permit B={2,3,8}

 

AxB={(1,2),(1,3),(1,8),(2,2),(2,3),(2,8),(3,2),(3,3),(3,8),(4,2),(4,3),(4,8)}

 

Count and you shall see there are 12 elements in set AXB.

 

Definition: For any set A, and any set B:

 

R is a binary relation from set A to set B if and only if R is a subset of set AXB.

 

Definition: For any set A:

 

R is a binary relation on set A if and only if R is a subset of AXA.

 

 

By one of the definitions above, the following is true:

 

For any set A: AXA ={(x,y)| x in A AND y in A}

 

The LHS is called the cartesian product of set A with itself.

 

Now, we need the definition of subset of a set.

 

Suppose that A denotes a set and B denotes a set.

 

Further suppose that the following statement is true:

 

If x is an element of A then x is an element of B, for any x.

 

Then, by definition, A is a subset of B.

 

If the converse is false, then A is a proper subset of B, and if the converse is true then A=B.

 

Definition: A Í B iff

 

1. A denotes a set.

2. B denotes a set.

3. If x is an element of A then x is an element of B, for any x.

 

I am going to write 3 above symbolically.

 

3. "x [xÎAÞxÎB].

 

A Í B is read "A is a subset of B"

 

Now, from here we can branch to the meaning of set theoretic equality, as well as the concept of a proper subset.

 

Suppose we are in a situation where we know that A denotes a set, and that B denotes a set, and that A is a subset of B.

 

There are now two remaining possibilities. Either Set A=Set B, or set A is a proper subset of Set B. And this just an issue of whether or not the converse follows. If the converse does follow, then A=B, and if the converse doesn't follow then A is a proper subset of B. In other words, if the converse is a true statement, then A=B, and if the converse is a false statement then A proper subset B.

 

Here is the statement that is the converse:

 

 

3. "x [xÎBÞxÎA].

 

So we have two more definitions:

 

Definition: A Ì B iff

 

1. A denotes a set.

2. B denotes a set.

3. "x [xÎAÞxÎB].

4. Ø"x [xÎBÞxÎA].

 

A Ì B is read "A is a proper subset of B"

 

Definition: Let A,B denote sets.

 

A=B if and only if

 

1. "x [xÎAÞxÎB].

2. "x [xÎBÞxÎA].

 

A=B is read "set A is equivalent to set B"

 

We can combine the two first order statements above into one, if we so choose:

 

Definition: Let A,B denote sets.

 

A=B if and only if "x [xÎAÛxÎB].

 

So now, I think theres enough above to show that defining a binary relation on a set A, as a subset of AXA leads to the concept of a set of all sets.

 

We have above:

 

Definition: For any set A:

 

R is a binary relation on set A if and only if R is a subset of AXA.

 

X is a binary relation hence:

 

Definition: For any set A:

 

X is a binary relation on set A if and only if X Í AXA.

 

We have above:

 

Definition: A Í B iff

 

1. A denotes a set.

2. B denotes a set.

3. "x [xÎAÞxÎB].

 

A Í B is read "A is a subset of B"

 

So, using the definitions above, it is necessarily true that:

 

 

For any set A:

X is a binary relation on set A if and only if

1. X denotes a set.

2. AXA denotes a set.

3. "x [xÎXÞxÎAXA].

 

Now, given that A is a set, if follows that AXA must denote a set, so we can shorten the above to:

 

For any set A:

X is a binary relation on set A if and only if

1. X denotes a set.

2. "x [xÎXÞxÎAXA].

 

Yesterday, I was concerned with whether or not the set A above, now denotes the set of sets.

 

Number 2 above seems funny. But again, its not my fault, i gave the "accepted" definitions.

 

Well I don't see a way to get to A={x|x is a set} from the above, and its bored me to tears so i'll leave it here.

 

Well i could look at one more thing. Write two above as:

 

2. "(x,y) [(x,y)ÎXÞ(x,y)ÎAXA].

 

Now, from a definition above, AXA is the set of ordered pairs (x,y) such that X in A, and Y in A.

 

That means that (x,y) in AXA if and only if (x in A AND y in A). So they are tautologically equivalent. So we can rewrite 2 above as:

 

2. "(x,y) [(x,y)ÎXÞ(xÎA AND yÎA)]

 

I think i should have used prenex normal form for this.

 

I'm going to leave things here, its going to give me a headache.

 

This is exactly why I couldn't stand axiomatic set theory.

[Johnny5]

This thread is supposed to be about the group axioms.

 

I just read that quaternions are a group, they are denoted by H, after Hamilton.

Also used is the symbol Q₈, but I think thats for Octonians, though Wolfram seems to imply otherwise.

 

Regardless, quaternions are a group, and specifically as matt says a division algebra.

[matt grime]

A group with respect to what operation Johnny. Trivially they are a group under addition since they are a vector space. They are not a group under multiplication.

[Johnny5]

A group with respect to what operation Johnny. Trivially they are a group under addition since they are a vector space. They are not a group under multiplication.

 

I know the answer to your question is "multiplication." I recently read that but I don't remember at which site, and ive looked at maybe 15-20 so far trying to find someone who makes sense.

 

Let me find the site, since you say addition.

 

 

At wolfram you will read that they are a group under multiplication.

 

Wolfram on quaternions

 

 

The quaternions ± 1, ±i, ±j, ±k, and form a non-Abelian group of order eight (with multiplication as the group operation).

 

So Matt, why did you say that they arent a group under multiplication?

 

Wolfram seems to imply that multiplication is the group operation.

[matt grime]

They (the quarternions as a division algebra as you called them yourself in this thread) are not a group under multiplication, just as the real numbers are not a group under multiplication. What is the mutliplicative inverse of 0? The non-zero elements are a group undermultiplication. The link you posted to is about Q_8 the quarternion group (of integral unit length quarternions) comes around from the fact that the elements +/-1, =/-i, +/-j and +/-k are a finite group of order 8 under multiplication.

 

You are confusing the Quarternionic group with the Quarternionic algebra (strictly speaking there are an infinitenumber of quarternionic algebras, this being the one we commonly talk about), this isn't that surprising, since they are both commonly called just quarternions. They are different. Only the second is a division algebra. The first is not an algebra. Why do I suspect you do not know what an algebra is?

[Johnny5]

You are confusing the Quarternionic group with the Quarternionic algebra (strictly speaking there are an infinitenumber of quarternionic algebras' date=' this being the one we commonly talk about), this isn't that surprising, since they are both commonly called just quarternions. They are different. Only the second is a division algebra. The first is not an algebra. Why do I suspect you do not know what an algebra is?[/quote']

 

Well i don't want to confuse one with the other, but that cannot happen since i dont know what either of them are.

 

Quaternionic group

Quaternionic algebra

 

What's the difference between them?

 

What kind of algebra are you thinking of? There are so many "things" called algebras, there is no one definition that would fit them all.

 

Regards

[matt grime]

So, in reading up on the quaternions you didn't bother to look up the definition? You have confused them, Johnny precisely because you don't know what either of them is.

 

1. An algebra is a ring that is a vector space (over a field) It is the definition of what an algebra is.

 

2. The quaternions are R^4 equipped with a product that satisfies the correct properties to be a ring and is thus an algebra.

 

We declare the basis to be given as 1,i,j,k with the rules you know so well: ij=k jk=i ki=j and ji=-k kj=-i ki=-j

 

Thus H is the set of all symbols a+bi+cj+dk where a,b,c,d are in R and mutliplication extended linearly. I've told you at least once before if not twice.

 

Q_8, the quarternion group is the set of quaternions with exactly one of a,b,c or d equal to plus or minus 1. It is a group with 8 elements.

 

You recall that you implied that the quaternionic people had made a mistake with the naming? Well, if you don't even know what they are how can you say such things?

[Johnny5]

So' date=' in reading up on the quaternions you didn't bother to look up the definition? You have confused them, Johnny precisely because you don't know what either of them is.

 

1. An algebra is a ring that is a vector space (over a field) It is the definition of what an algebra is.

 

2. The quaternions are R^4 equipped with a product that satisfies the correct properties to be a ring and is thus an algebra.

 

We declare the basis to be given as 1,i,j,k with the rules you know so well: ij=k jk=i ki=j and ji=-k kj=-i ki=-j

 

Thus H is the set of all symbols a+bi+cj+dk where a,b,c,d are in R and mutliplication extended linearly. I've told you at least once before if not twice.

 

Q_8, the quarternion group is the set of quaternions with exactly one of a,b,c or d equal to plus or minus 1. It is a group with 8 elements.

 

You recall that you implied that the quaternionic people had made a mistake with the naming? Well, if you don't even know what they are how can you say such things?[/quote']

 

 

It's ok, I'm beginning to get a handle on it (quaternions).

 

An algebra is a ring, which is a vector space over a field.

 

What is a ring again?

 

 

And a couple of important questions.

 

Firstly, what is R^4??? I am a person who holds that space is three dimensional, so what does the fourth dimension of quaternionic algebra represent. From what I have been able to gather, from reading multiple sources, is that the extra dimension happens to correspond to a rotation. An angle.

 

And i have one other question. I was running through the mathematics of the product of two quaternions, and something bothered me. I was trying to multiply two quaternions together, and get the cross product and the dot product. I have a question on it when you get a chance.

 

I can explain the problem.

 

I was able to generate the cross product, using

 

ij=k, ji=-k

ik=j, ki=-j

jk=i, kj=-i

 

But, for ii i got 0, instead of -1, for jj i got 0, instead of -1, and for kk i got zero instead of -1. So i am sure i made some kind of mistake, but i dont know where, if you could show me just that alone, i would be grateful.

 

Regards

[Dave]

What is a ring again?

 

A ring is a set R with two binary operations, + and *. R must contain a zero element (but not necessarily an identity element); + must satisfy associativity, additive inverse and commutativity; * must be associative. On top of this, you have the distributivity laws, a(b+c) = ab + ac, (b+c)a = ba + ca.

 

Firstly, what is R^4???

 

[math]\R^4 = \{ (x_1, \dots, x_4) \ | \ x_1, \dots, x_4 \in \R \}[/math]. Or, alternatively, [math]\R^4 = \R \times \R \times \R \times \R[/math].

 

I am a person who holds that space is three dimensional, so what does the fourth dimension of quaternionic algebra represent.

 

It's not supposed to represent anything. [math]\R^4[/math] is a Cartesian product of sets and as such you're not supposed to infer whether this implies any physical representation - which it doesn't, I might add. I don't want to get into the entire thing about things in maths been represented in physical quantites because I don't want to ruin another thread that hasn't been that bad so far.

 

But, for ii i got 0, instead of -1, for jj i got 0, instead of 1, and for kk i got zero instead of one. So i am sure i made some kind of mistake, but i dont know where, if you could show me just that alone, i would be grateful.

 

Well, unless you show the working we can't point out the error

[matt grime]

ixi is indeed zero for the cross product, but i*i is not ixi is it?

 

And who cares what you hold true about space. R^n is just a set with extra structure as Dave rightly points out. It may or may not correspond to anything that you hold to be real, but that is completely immaterial.

 

Your comment about "beginning to get a handle on it" was quite funny. It reminds me of something Tom Koerner said once: a Fellow at Trinity Hall, who knows nothing of mathematics, commented to him that he'd seen the panorama documentary on Andrew Wiles and thought he almost understood the proof of Fermat's Last Theorem.

[Johnny5]

ixi is indeed zero for the cross product' date=' but i*i is not ixi is it?

[/quote']

 

OK yeah i thought that too, no i*i isnt iXi but...

 

In order to get the other relations, i assumed that it was a cross product.

 

Here, this is for Dave too.

 

Quaternion 1: (s1,x1,y1,z1)

Quaternion 2: (s2,x2,y2,z2)

 

All components elements of real number system.

 

Examine the following special case:

 

s1=s2=0

 

So we have:

 

Quaternion 1: (0,x1,y1,z1)

Quaternion 2: (0,x2,y2,z2)

 

Now, multiply two quaternions together, and interpret the multiplication as the vector cross product.

 

[math] (x_1 \hat i + y_1 \hat j + z_1 \hat k) \times (x_2 \hat i + y_2 \hat j + z_2 \hat k) [/math]

 

So we have:

 

[math] x_1 \hat i \times x_2 \hat i + x_1 \hat i \times y_2 \hat j + x_1 \hat i \times z_2 \hat k + [/math]

 

[math] y_1 \hat j \times x_2 \hat i + y_1 \hat j \times y_2 \hat j + y_1 \hat j \times z_2 \hat k + [/math]

 

[math] z_1 \hat k \times x_2 \hat i + z_1 \hat k \times y_2 \hat j + z_1 \hat k \times z_2 \hat k [/math]

 

As you can see, there are nine terms. I think thats called a dyadic tensor, but i dont remember. Anyhow...

 

Assume that we have associativity of scalars, and associativity of vectors.

 

Furthermore assume commutativity of vectors times scalars. Hence we can write the above as:

 

[math] x_1 x_2 \hat i \times \hat i + x_1 y_2 \hat i \times \hat j + x_1 z_2 \hat i \times \hat k + [/math]

 

[math] y_1 x_2 \hat j \times \hat i + y_1 y_2 \hat j \times \hat j + y_1 z_2 \hat j \times \hat k + [/math]

 

[math] z_1 x_2 \hat k \times \hat i + z_1 y_2 \hat k \times \hat j + z_1 z_2\hat k \times \hat k [/math]

 

Interpreting X as vector cross product the above is equivalent to:

 

[math] x_1 x_2 \hat i \times \hat i + x_1 y_2 \hat k + x_1 z_2 \hat j + [/math]

 

[math] -y_1 x_2 \hat k + y_1 y_2 \hat j \times \hat j + y_1 z_2 \hat i + [/math]

 

[math] -z_1 x_2 \hat j -z_1 y_2 \hat i + z_1 z_2\hat k \times \hat k [/math]

 

The six terms i just changed, correspond to the cross product of two vectors.

 

Thus, the above is equivalent to:

 

[math] x_1 x_2 \hat i \times \hat i+ y_1 y_2 \hat j \times \hat j + z_1 z_2\hat k \times \hat k + \mathbf{V_1} \times \mathbf{V_2} [/math]

 

Where

 

V1 = <x1,y1,z1> = x1i + y1j + z1k

V2 = <x2,y2,z2> = x2i + y2j + z2k

 

The problem that I now have though, is that iXi=0, jXj=0, and kXk=0, but from various sources, I should get the negative of the scalar dot product, plus the vector cross product, not simply the vector cross product. So they cannot be equal to zero, they must satisfy

 

iXi=-1=jXj=kXk

 

but they do not.

 

 

PS: And Dave, that has to be the quickest exposition on Rings I've ever seen. That was perfect. Watch how it keeps.