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Math Help - Show centralizer of a = centralizer of a^-1

  1. #1
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    Show centralizer of a = centralizer of a^-1

    Help. I feel like i'm over complicating this in my head.

    Let G be a group, and let a be in G. Prove that C(a)=C(a^-1).
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  2. #2
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    Quote Originally Posted by baseballmath View Post
    Help. I feel like i'm over complicating this in my head.

    Let G be a group, and let a be in G. Prove that C(a)=C(a^-1).
    1)Show if x\in C(a^{-1}) then x\in C(a)
    2)Show if x\in C(a) then x\in C(a^{-1})
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  3. #3
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    I'm struggling with getting to show that. It's given that ag=ga and a^1g=ga^1.

    I tried to go with
    a^1 g a
    = g a^1 a
    = g e

    and: a g a^1
    = g a a^1
    = g e
    so a a^1 = a^1 a But I feel I'm making assumptions somewhere so the proof doesn't hold. I'm so lost! Thanks for your reply, it makes sense, I'm just not sure how to show what you're saying...
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  4. #4
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    I do one direction.
    Let x\in C(a).
    Therefore, x^{-1} \in C(a) because C(a) is a subgroup.
    By definition ax^{-1} = x^{-1}a.
    Thus, (ax^{-1})^{-1} = (x^{-1}a)^{-1}.
    Thus, xa^{-1} = a^{-1}x.
    By definition x\in C(a^{-1}).
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