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Math Help - Multiplicative Group Proof by Using Given Properties (Not That Hard, I suppose)

  1. #1
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    Multiplicative Group Proof by Using Given Properties (Not That Hard, I suppose)

    Let G be a multiplicative group (not Abel's group, so it's non-commutative) and a and b are elements that belong to it. Also, a = 1 and b = a^-1 * b * a. Knowing this, show that b^5 = 1.

    From a = 1, I can see derive that a = a^-1. I have also derived that b = aba and
    b = aba, but I don't know if I'm on the right track. I've tried other ways too, but I always seem to end up where I started. I suppose this task isn't that hard, but I just need a little extra push to be able to solve it. I've been thinking if there's a way to state b in form b = a^n. This would solve probably everything, since a^(2n)=1.

    Any help would be appreciated. Thanks in advance!
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  2. #2
    MHF Contributor Amer's Avatar
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    b^5 = b^2 b^3

    b^3 = a b^2 a from this a b^3 = b^2 a

    b^5 = b^2 (a b^2 a )
    b^5 = b^2 a (b^2 a )

    can u continue ?
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  3. #3
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    Ah! So the key was to replace b^n terms with b^(n-1) terms until there were no b terms left. The final phases I got were

    (ba)(ba) = (bab)ba = baba = (bab)a = aa = a = 1.

    Here I used the following equivalences that I got from further manipulating b = aba:
    bab = ba
    bab = a

    I suppose this is correct. Thanks a lot for the hint! =)
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  4. #4
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    Quote Originally Posted by Koaske View Post
    Ah! So the key was to replace b^n terms with b^(n-1) terms until there were no b terms left. The final phases I got were

    (ba)(ba) = (bab)ba = baba = (bab)a = aa = a = 1.

    Here I used the following equivalences that I got from further manipulating b = aba:

    bab = ba

    I can't see from where the above follows, and from this precisely follows that b^5=1

    Tonio


    bab = a

    I suppose this is correct. Thanks a lot for the hint! =)
    .
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  5. #5
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    I can't see from where the above follows, and from this precisely follows that

    Tonio
    I got bab = ba in the following way:

    b = aba <=> bb = aba <=> (aba)b = aba (replaced b = aba)
    <=> abab = aba <=> bab = ba (multiplied from left with a, then with b^-1)

    In this, I also used the following information:
    G is a multiplicative group, so there has to exist inverse b^-1 so that b^-1 * b= e.
    We already knew that a^-1 = a and that a = 1, so it must be that the neutral element e = 1.
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