Identity element

Originally posted by DeepThought
To be an identity element it [b]has to have the property that e*x = x for each element x in the group. Just one isn't enough. That is the problem I was trying to solve and why I was struggling. I should have spotted there was a problem since commutativity implies there is a unique identity element.

Besides my earlier categorization is sufficie ...[text shortened]... 1 and 1 with the normal rules of multiplication) consists of an identity and a type III element.[/b]

To be an identity element it has to have the property that e*x = x for each element x in the group

You're stating the definition of the identity (not an identity element) element in group theory. The thing is this is not a group. The structure we're looking at actually has a mathematical name (that I won't give away, so that people don't look it up and spoil the fun) and in this context there is such a thing as multiple identities. Given you're previous description of your PhD. I'm pretty sure that you've encountered some of this stuff in your study of algebras.

So please once and for all forget your group theory mentality in this simple example because some of it just doesn't apply here.

Suppose the set S = {a, b, c, ..., k, l}, and is closed under * which is associative and commutative.
a*b*c*···*k*l is the product of all the elements and is in the set (call it e), then we have:
e = e*(a*b*··*d*f*···*k*l) = e*z

if
a*b*c*···*k*l=e (by your definition)

then in general you can't affirm without proof that

e = e*(a*b*··*d*f*···*k*l)

Then you go around and define

z=(a*b*··*d*f*···*k*l)

when first you said that

(a*b*··*d*f*···*k*l) =e

so that makes it z=e

This means that z acts as an identity with respect to all elements obtainable by multiplying e by each member of the set...

;

Apparently you defined z to be equal to e and you also apparently wrongly assumed that e=e*(a*b*··*d*f*···*k*l). That's why you're getting this wrong result

...but this is not guaranteed to span the set we need an extra axiom for that.

Here I really don't follow your argument. The problem statement is clear on the fact that this set is closed under the * operation so why do you need to span the whole set?

PS: You're always picking new letters and this makes your argument very hard to follow and I think that this may be the root cause of your confusions (together with the fact that you're implicitly assuming that this structure is a group and it isn't)
PPS: Can you find a mistake with Agerg's proof or any inconsistency with the problem stating (taking the fact that I initially omitted the fact that * is also associative)
PPS: After we end up tidying up your arguments I'll follow through my intended sequence for the series of "posts" I'm thinking about.

Originally posted by adam warlock

To be an identity element it [b]has to have the property that e*x = x for each element x in the group

You're stating the definition of the identity (not an identity element) element in group theory. The thing is this is not a group. The structure we're looking at actually has a mathematical name (that I won't give away, so t ...[text shortened]... arguments I'll follow through my intended sequence for the series of "posts" I'm thinking about.[/b]

This could be author dependent, but in any usage I've heard of an identity element has to be an identity for each element of a set whether or not it is a group. There can be more than one identity. This example is from the Wikipedia page titled identity element:

S = {e,f} with an operation * which is not commutative.
e = e*e = f*e
f = e*f = f*f
There is no right identity, both e and f are left identities.

The z = e thing at the end was the result of some confusion on my part about what the result of the product was. The rest is all correct - the problem with Agerg's notation is that it's a pain to type in. I wasn't disagreeing with him.

Choose a (yes it's a new letter, but this makes things clearer as e is often used to denote the identity) to be the name of the element which is the product of all elements, since we have a commutative and associative operation we can always reorder the product so a is at the start. The product is then:

a = a*b*c*···*g*h
let b*c*···*g*h = z (this is what I'd intended you to read into my earlier post)

Then a = a*z
The above was just to establish notation my point was the following, let x be any element of the magma, then:

y = x*a = x*a*z = y*z

So z is acts as an identity with respect to all elements y which can be obtained by multiplying a by any element of the set, this operation does not necessarily generate the entire set. So we can't call z an identity element yet (by my definition of identity). We also haven't ruled out the possibility that there is more than one identity (in my sense), or no identity. This requires an extra axiom.

If z is an identity element (by my definition of identity) then we can find inverses, suppose we want the inverse for b:

z = b*c*···*g*h
z = b*(c*d*···*g*h)
=> inv(b) = c*d*···*g*h

This trick will work for all elements of the set except for a, since the commutativity and associativity axioms allow us to shuffle elements around to our hearts content. Interestingly we have trouble finding an inverse for a. For a concrete example consider {1, -1, i, -i}, with the normal rule for multiplication (i*i = -1):
1 * -1 * i * -i = -1
so z = 1*i*-i = 1 (as we'd hope)
inv(i) = 1*-i
inv(-i) = 1*i
inv(1) = i*-i
clearly inv(-1) = -1 but we can't see that from the abstract considerations above.

Incidentally for Z_3 {1, exp(2*i*pi/3), exp(-2*i*pi/3)} the product of the elements is the identity element.

This implies to me that the product of all the elements (a) is self-inverse.

Originally posted by DeepThought
Choose a (yes it's a new letter, but this makes things clearer as e is often used to denote the identity) to be the name of the element which is the product of all elements, since we have a commutative and associative operation we can always reorder the product so a is at the start. The product is then:

a = a*b*c*···*g*h
let b*c*···*g*h = z (this is ...[text shortened]... dentity element.

This implies to me that the product of all the elements (a) is self-inverse.

a = a*b*c*···*g*h
let b*c*···*g*h = z (this is what I'd intended you to read into my earlier post)

Then a = a*z
The above was just to establish notation my point was the following, let x be any element of the magma...

Ok so you've presented Agerg's proof with a different notation.

So z is acts as an identity with respect to all elements y which can be obtained by multiplying a by any element of the set, this operation does not necessarily generate the entire set. So we can't call z an identity element yet (by my definition of identity). We also haven't ruled out the possibility that there is more than one identity (in my sense), or no identity. This requires an extra axiom.

Let us clearly fix terminology from now on so that we're all on the same page:

1 - An identity is something that acts on an element of the set and the end result is the same element of the set.
2 - The identity is element of the set that acts on all elements of the set and the end result is the element of the set we started with.

Since * is commutative we need no to distinguish between a left and a right identity.

Now what I still don't get in your argument is the point that you make by saying that "this operation does not necessarily generate the entire set". What's wrong if the operation does not generate the entire set (I'm not saying it doesn't, I'm trying to understand what's your point if it doesn't)?

Edit: Stop spoiling the future of this series man! 😛😛😛

Originally posted by adam warlock

a = a*b*c*···*g*h
let b*c*···*g*h = z (this is what I'd intended you to read into my earlier post)

Then a = a*z
The above was just to establish notation my point was the following, let x be any element of the magma...

Ok so you've presented Agerg's proof with a different notation.

[quote]So z is acts as an identity with respect t ...[text shortened]... d what's your point if it doesn't)?

Edit: Stop spoiling the future of this series man! 😛😛😛

Ok. I'll accept your version of identity - I'd regard it as an abuse of notation (in the sense that it's allowed when it's clear what it means) - but we should then use the term universal identity to mean an identity in my sense (unless it's obvious which we mean). I think if an element is both a left and a right universal identity then it's unique, so since the set under consideration is commutative there's no scope for confusion.

The point I was making is that in one of your earlier posts you seemed to be saying that we could show that z (as I was calling it) could be shown to be an identity w.r.t. all elements of the set. If a*x generates each element of the set we can show that. Otherwise we need to add an axiom - unless I've missed a trick.

Originally posted by DeepThought
Ok. I'll accept your version of identity - I'd regard it as an abuse of notation (in the sense that it's allowed when it's clear what it means) - but we should then use the term universal identity to mean an identity in my sense (unless it's obvious which we mean). I think if an element is both a left and a right universal identity then it's unique, so ...[text shortened]... ent of the set we can show that. Otherwise we need to add an axiom - unless I've missed a trick.

Ok. I'll accept your version of identity - I'd regard it as an abuse of notation

Actually it isn't an abuse of notation but it is stretching a concept out of our comfort zone.

In the link you provided about the identity concept you can check the Properties section and they have the following text (my emphasis):

As the last example (a semigroup) shows, it is possible for (S, &lowast😉 to have several left identities

When you dealing with structures that aren't groups you can have various identities that are different from each other. And that's what's fun about it!

Usually the distinction that people make is between the expressions "an identity" and "the identity" or just "identity".

The point I was making is that in one of your earlier posts you seemed to be saying that we could show that z (as I was calling it) could be shown to be an identity w.r.t. all elements of the set.

Ok now I get it. Sorry if I came across wrong but I only was saying that it was an identity for a specific element of the set and not all elements of the set.

I thought of another little example last night when I couldn't sleep. Consider the two element set {a, e} with a*a = a, a*e = e, e*e = e. e is a perfectly good universal identity element, despite that a has no inverse, so it's not a group.

Also it's possible to make some progress classifying these things. Take my type III elements. If we have an N element set {a1, a2, ..., aN} with an*an = a{n+1}, except for aN*aN = aN, the product of the elements p is:
p = a1*a2*···*aN,
p*p = a1*a1*a2*a2*···*aN*aN = a2*a3*···*aN
eventually repeatedly squaring we get:
p^{2(N-1)} = aN,
This means we can identify p, it must be a1 as it is the only element that needs squaring that many times to reduce to aN. So we have:
a2*a3*···*aN, as an identity w.r.t. a1. But repeating the above for this element gives:
a2 = a2*a3*···*aN
This means that if we are going to keep associativity (see my three element example from when I was trying to build a counterexample above) and commutativity then we have:
a2 = a3 = a4 = ··· = aN
So chains of elements like this can only be one link long, if they aren't branching. c.f. Z_4 = {-i, i, -1, 1} under the squaring operation regarding each element as a dot and joining the dots iff an element squares to give the next element then it looks like the letter Y, with 1 at the bottom, -1 at the fork and ±i at the top of each prong. A V won't work if a*a = e and b*b = e, e*e = e then:
p = a*b*e
p*p = e,
If p = a then b*e = inv(a). But a*a = e, so a = inv(a) = b*e. If e is to be a universal identity we end up with a = b.

Of course the way I have the chain of elements set up a1*a1 = a2, a2*a2 = a3 etc. we have:
a1 = a
a2 = a²
a3 = a^4
a4 = a^8
etc.
there are clearly elements missing when it's written like this. So I'm going to venture the guess that any commutative group is a sub-group of U(1) the circle group, or of cartesian products of it U(1)̣×U(1)×···×U(1).

Originally posted by DeepThought
Of course the way I have the chain of elements set up a1*a1 = a2, a2*a2 = a3 etc. we have:
a1 = a
a2 = a²
a3 = a^4
a4 = a^8
etc.
there are clearly elements missing when it's written like this. So I'm going to venture the guess that any commutative group is a sub-group of U(1) the circle group, or of cartesian products of it U(1)̣×U(1)×···×U(1).

This is a total aside but I thought you all might be interested:

http://phys.org/news/2014-12-xinwen-zhu-discusses-theory-mathematics.html#nRlv

He is looking for a unifying field in maths.