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Proof Using Def. of Groups

  • Thread starter wubie
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wubie

[SOLVED] Proof Using Def. of Groups

Hello,

This is my question:


Let G be a group.

i) Let x and y be elements of G. Prove that (xy)2 = x2y2 iff xy = yx. (Hint: Use the definition g2 = gg).

ii) Using part (i) prove that if g2 = u (the unit element) for all g which is an element of G, then G is abelian.


Now I BELIEVE that I have properly proved the part (i) of the question. But I am not sure how to proceed with part (ii). In fact, the second part question makes me wonder if I did part (i) correctly.

I know that the definition of abelian is:

For every x and y which are elements of G, a group G with the property x o y = y o x is called abelian ( or commutative). To rephrase, I would think this is the same as F(y,x) = F(x,y).

Now I am not sure what the definition would be in context of the question. Is the question saying,

Proposition: If g2 = u then xy = yx?
Is that the proposition that I am supposed to prove? And if that is the case, I am still not sure how to use the nfo g2 = u. How does it apply to the relation in part (i)?

In the case of part (i) would this be it?

x o y = x2y2. Then

x o u = x = u o x --> x2 * u = x = u * x2? (In which case 1 would be the identity element. Correct?).


Any help/clarification is appreciated. Thankyou.
 

Tom Mattson

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Originally posted by wubie
Proposition: If g2 = u then xy = yx?

Is that the proposition that I am supposed to prove?
Yes.

And if that is the case, I am still not sure how to use the nfo g2 = u. How does it apply to the relation in part (i)?
Consider the product fg, where f and g are both in G. Since the group is closed, fg is also in g. We then must have, according to the info they gave (gg=u):

(fg)(fg)=u

Right multiply by gf:

(fg)(fg)(gf)=u(gf)

Can you take it from there?

In the case of part (i) would this be it?

x o y = x2y2.
That should be (x o y)o(x o y)=(x o x)o(y o y).
 

wubie

Consider the product fg, where f and g are both in G.
Right. Then if f and g are elements of G then by definition of a group, f o g is an element of G.
We then must have, according to the info they gave (gg=u)
Where u is an element of G by definition of a group and also by the aforementioned property. Correct?
(fg)(fg)=u
Right multiply by gf:

(fg)(fg)(gf)=u(gf)

Can you take it from there?
Yes. I can see it now. Perhaps you did too much in this step.
That should be (x o y)o(x o y)=(x o x)o(y o y).
OH! That helps! It's like a composition function right? For instance,

q = F(y,x) = x * y and
G(q)= G (F(y,x)) = q * q = F(y,x) * F(y,x) = (x*y)*(x*y).

This would be the same for the right side of the equation yes?

I will be back later and post the proof.

Thankyou Tom.
 

Tom Mattson

Staff Emeritus
Science Advisor
Gold Member
5,475
20
Originally posted by wubie
Right. Then if f and g are elements of G then by definition of a group, f o g is an element of G.

Where u is an element of G by definition of a group and also by the aforementioned property. Correct?
Yes; the group must have an identity element.

OH! That helps! It's like a composition function right?


For instance,

q = F(y,x) = x * y and
G(q)= G (F(y,x)) = q * q = F(y,x) * F(y,x) = (x*y)*(x*y).
Right.

This would be the same for the right side of the equation yes?
In the language of your function F(x,y), you have:

F(x,y) o F(x,y)=F(x,x) o F(y,y)
 

wubie

In the language of your function F(x,y), you have:

F(x,y) o F(x,y)=F(x,x) o F(y,y)
Right.Right. My mistake. Thanks alot Tom. Your help has been great increasing my understanding of the subject.

Cheers.
 

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