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Math Help - Algebra in Groups

  1. #1
    No one in Particular VonNemo19's Avatar
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    Detroit, MI
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    Algebra in Groups

    Hi.

    solve simultaneously:

    x^2=a^2

    x^5=e (e is the identity element in G)

    Um...

    x=a^(5k) k in the positive integers?

    Easy guys, I'm a beginner.

    Thanks.
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  2. #2
    MHF Contributor

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    Re: Algebra in Groups

    is the idea that we have to solve for x in terms of a?

    if so, then e = x^5 = (x^2)(x^2)x = (a^2)(a^2)x = (a^4)x so therefore:

    (a^-4) = (a^-4)e = (a^-4)(a^4x) = (a^-4a^4)x = ex = x.
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