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Proof Using Def. of Groups

  1. Oct 9, 2003 #1
    [SOLVED] Proof Using Def. of Groups


    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,

    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.
  2. jcsd
  3. Oct 9, 2003 #2

    Tom Mattson

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    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):


    Right multiply by gf:


    Can you take it from there?

    That should be (x o y)o(x o y)=(x o x)o(y o y).
  4. Oct 9, 2003 #3
    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. I can see it now. Perhaps you did too much in this step.
    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.
  5. Oct 9, 2003 #4

    Tom Mattson

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    Yes; the group must have an identity element.


    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)
  6. Oct 9, 2003 #5
    Right.Right. My mistake. Thanks alot Tom. Your help has been great increasing my understanding of the subject.

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