I was re-reading unit GTB1 tonight, of the Open University course M208 Pure Mathematics.
I just thought I would note down something that I have noticed, but which I have only just realized, on reviewing the material, this year.
Just a bit of background; M208 was the first pure mathematics course that I completed, and part of it contained sections on Group Theory. As any diligent group theorist would do, I began by learning the Axoims that lead to the definition of a 'Group', from a set of any given elements.
Now, one of those axioms is the property of associativity. That is:
For all g1, g2, g3 elements of G,
g1 o (g2 o g3) = (g1 o g2) o g3.
(the numbers are supposed to be subscript, but my Latex editor is currently 'up the Shoot'.)
Whilst I was studying the course, I had assumed that a group probably needed three elements in a set, in order to meet this axiom's criteria.
I say I assumed this, however, to be honest, I had never actually given it any real thought.
So, on revision of this material in 'slow time'; this is the sort of question, that seems to be popping into my head. A promising sign, in my eyes.
So, I glanced at the issue tonight and, of course, I quickly concluded that the axioms do not state that you need three elements in a set, in order to meet their requirements.
For example, if I have a set {1}; this could be a group under multiplication. Why? Well, it meets the axioms of Closure, Identity, Inverses and Associativity; where g1, g2 and g3 are all the set element {1}. There are no axioms that say that the elements in a set must be different, or that the elements must be distinct.
Why? Well, the axioms do not go into numbers of elements, or other topics, for that matter. The way that they are written, means that they do not assume that there is more than one element or even that the elements actually exist.
Having said that the Axioms certainly do not state, that there should be a none existent element, or that there should be a group that exists, which contains less than three elements.
So, the axioms certainly don't prove that there is a set of less than three elements that can form a group.
Logical, but confusing, I think.
The only group with 1 element would be the identity element.
ReplyDeleteSimilarly the only possible group with 2 elements is one with identity element and another element.
However associativity is an operation not an element. You can apply group multiplication as many times as you like and associativity provides a constraint on the operation
namely that (e o a )o e for example is the same as eo(aoe) Not that that is particularly exciting.
It just corresponds to repeated reflections, its always good to go back and consolidate stuff. I never really got the hang of the counting theorem as I was more concerned with analysis.
Best wishes Chris
Theorem
ReplyDeleteIn every group, the identity, and only the identity, has order 1.
Proof
Let G be a group with identity e.
e1=e.
∀a∈G:a≠e:a1=a≠e.
http://www.proofwiki.org/wiki/Identity_Only_Group_Element_Order_1
Regards
DMC