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Math Help - Group Problem

  1. #1
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    Group Problem

    Prove that if (ab)^{2}=a^{2}b^{2} in a group G, then ab=ba.

    My proof so far:

    Now (ab)^{2}=a^{2}b^{2} for each a,b \in G,

    so a^{2}b^{2}=a^{2}b^{2}

    implies a^{2}b^{2}=b^{2}a^{2}

    then, there exist a^{-1},b^{-1} in G

    (a^-1)(a^2b^2)(b^-1) = (a^-1)(b^2a^2)(b^-1)

    ab=ba.

    Am I allow to switch the a and b like that?
    Last edited by tttcomrader; August 25th 2007 at 03:27 PM.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Prove that if ab^{2}=a^{2}b^{2} in a group G, then ab=ba.
    Have you written the problem correctly?
    Or should it be (ab)^{2}=a^{2}b^{2}???

    If you do mean that then:
     \left( {ab} \right)^2  = a^2 b^2
    \left( {ab} \right)^2  = \left( {ab} \right)\left( {ab} \right) = a^2 b^2
     a^{ - 1} \left( {ab} \right)\left( {ab} \right)b^{ - 1}  = a^{ - 1} \left( {a^2 b^2 } \right)b^{ - 1}
     ba = ab
    Last edited by Plato; August 25th 2007 at 03:26 PM.
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  3. #3
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    Oops, yeah, that is right.

    <br />
(ab)^{2}=a^{2}b^{2}<br />
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  4. #4
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    Please see my edit.
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  5. #5
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    Thank you, sir, that is really easy, I really should have gotten it myself.
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