Group Theory Help

[ed84c]

Im currently doing some teach-yourself-maths in preperation for exams in June and am on the last chapter, Group theory, which I seem to be struggling with.

 

Here's the question im stuck with;

 

"G is a group and D(G) is the set of ordered pairs of elements (g,e) where is an element of G and e = +/- 1. Show that d(G) together with the binary operation (g,e) @ (h,d) = (gh^e,ed) is a group if and only if G is Abelian ((h,d) denotes another element of D(G))" (@ is a general binary operation)

 

Unfortunately the example in the book was not particularly helpful so I have no real idea how to tackle this.

 

What I do know;

The 4 axioms- closed, associativity, identity, inverse. So I'm assuming by the question 1 (or more) of these only applies if the group is ableian (i.e. has commutativity).

 

So ive done;

 

1. Identity exists- (1,1)
   
2. Inverse exists- (1/h,d) (if and only if d is +/- 1?)
   

 

Is this the right way to do things? If so where do I go next (for closed & associativity)?

[timo]

- You should add and write out the proofs for the identity and the inverse, e.g. (a.x) * (1,1) = (a,x) = (1,1)*(a,x) for the neutral element. You did not state what (1/h,d) is supposed to be the inverse of. Again: Do and -if you want feedback- post the full proof, not the result (the result that D is a group exactly if G is abelian is already given in the question, anyways).

- Closure is the simplest of all the steps, so try that next.

[ed84c]

Ok cheers;

 

(1/h,d) is the inverse to (gh^e,ed) if e is 1

 

How do I prove if something is closed or not?

 

From this- if I say g,h,d are integers, e is -1 then (gh^e) is not an integer- does this prove it isnt closed?!

[timo]

(1/h,d) is the inverse to (gh^e,ed) if e is 1

You want to show that every (a,x) has an inverse (A,X), i.e. (a,x)*(A,X) = 1 = (A,X)*(a,x). It does (usually) not help you to find inverses for arcane combinations of elements ( (gh^e,ed) rather than a simple (g,e) ) - you'd have to proof that all elements of D can be written as such a combination. It is not sufficient to show that all (a,+1) have inverses; the (a,-1) need to have inverses, too.

 

If e=1 then you are saying that (1/h,d) is the inverse to (gh,d). That's wrong for e.g. h=1, d=1, g!=1: (1,1)*(g,1) = (1*g,1) = (g,1) [math]\neq [/math] 1. Sidenote: 1 refers to the neutral element of the appropriate group (D, G or {+1,-1; *}).

 

How do I prove if something is closed or not?

You show that (a,x)*(b,y) is an element of D for all (a,x) and (b,y) in D. In this case the closure deduces directly from G and {+1,-1; *} being groups (i.e. also being closed).

 

From this- if I say g,h,d are integers, e is -1 then (gh^e) is not an integer- does this prove it isnt closed?!

Not if "it" is D. Integers are a group under addition not under multiplication (there is no multiplicative inverse of 2, for example) so integers under multiplication do not qualify as G. You seem to be thinking of multiplication in your example.

 

You seem completely confused about the topic. For your own clarification, please try to construct complete statements and do not just throw in results or guesses. It is almost impossible to understand what you are trying to say.

 

Remarks (possibly helpful; you seem to have problems understanding what the notations in the question mean):

- [math]a \, \in \, G: \ a^1 = a; a^{-1}[/math] means the inverse of a. 1 is the identity of G.

- Sample proof for the identity being (1,1): [math] (1,1)(a,x) = (1a^1,1x) = (a,x), \ (a,x)(1,1) = (a1^x,x1) = (a1,x) = (a,x)[/math], where [math]1^{-1}=1[/math] has been used in the 2nd equation (proof: 1*1=1=1*1 => 1^{-1} = 1).

- Be very careful about the notations. They are not what you are probably used to. If G was the integers under addition then 1*x, x*y and x^{-1} (for x,y in G) would read differently in the notation you are used from school math where addition is written with a "+". Namely 0+x, x+y and -x, respectively.

[ed84c]

OK thanks for the help- I'm going to back track a bit to try and get a fuller understanding of the topic and then come back to this.

 

 

Cheers anyway,

Ed

[DavyJonesLoquet]

Group set theory only is used for axionic compouding since we cannot say that data points in the inverse properties of logarythmic reduction set in communitive properties of co-tangent fields and thier super and giant set features.