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Math Help - Show G is not necessarily abelian

  1. #1
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    Show G is not necessarily abelian

    Let a,b be elements of a group G. Given that (ab)^i = (a^i)(b^i) for two consecutive integers i. Show that G is not necessarily abelian. I'm am unable to find an counter example to show this.
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  2. #2
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    Re: Show G is not necessarily abelian

    D_4?
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    Super Member ILikeSerena's Avatar
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    Re: Show G is not necessarily abelian

    Quote Originally Posted by wellfed View Post
    Let a,b be elements of a group G. Given that (ab)^i = (a^i)(b^i) for two consecutive integers i. Show that G is not necessarily abelian. I'm am unable to find an counter example to show this.
    Hi wellfed!

    Well, let's start with the smallest group that is not abelian.
    That is S3, the permutations of 3 elements.

    Now consider which values "i" are interesting.
    For instance, i=1 is okay, since you'll get a trivial identity.
    But for instance, i=2 is a problem, since abelian follows immediately (can you show that?).

    What are other interesting values for "i" in the sense that you can say something immediately (as related tot S3)?
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    Re: Show G is not necessarily abelian

    Quote Originally Posted by ILikeSerena View Post
    Hi wellfed!

    Well, let's start with the smallest group that is not abelian.
    That is S3, the permutations of 3 elements.

    Now consider which values "i" are interesting.
    For instance, i=1 is okay, since you'll get a trivial identity.
    But for instance, i=2 is a problem, since abelian follows immediately (can you show that?).

    What are other interesting values for "i" in the sense that you can say something immediately (as related tot S3)?
    Hi ILikeSerena,

    Thanks for the hint. I guess I simply didn't have the patience to examine the S3 carefully before.
    I found that (2 3)^6(1 2)^6=[(2 3)(1 2)]^6 = e , (2 3)^7(1 2)^7=[(2 3)(1 2)]^7 = (1 2 3) , and (2 3)(1 2) ne (1 2)(2 3).
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  5. #5
    Super Member ILikeSerena's Avatar
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    Re: Show G is not necessarily abelian

    Yep. Patience helps. ;-)

    Did you notice that you've selected the order of the group, which is equivalent to i=0?
    And also i=7, which is equivalent to i=1, since 6 is the order of the group?
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