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Math Help - Abstract Algebra

  1. #1
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    Abstract Algebra

    A little help needed. Prove: C(bab^-1)=bC(a)b^-1. I know that the centralizer states that every element a commutes with a. so can i say by definition of a centralizer b and its inverse doesnt commute therefore the right side is true? Thanks!!!
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  2. #2
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    I didn't quite understand what you meant, but you should solve this using the definition of centralizers
    C(b^{-1}ab)=\{g \in G | gb^{-1}ab = b^{-1}abg \} = \{g \in G | bgb^{-1}abb^{-1} = bb^{-1}abgb^{-1} \} = \{g \in G | bgb^{-1}a = abgb^{-1} \} = \{g \in G | bgb^{-1} \in C(a) \} = \{g \in G | g \in b^{-1}C(a)b \} = b^{-1}C(a)b
    so you get C(b^{-1}ab) = b^{-1}C(a)b
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  3. #3
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    Quote Originally Posted by GreenandGold View Post
    A little help needed. Prove: C(bab^-1)=bC(a)b^-1.
    Just show C(bab^{-1})\subseteq bC(a)b^{-1} and bC(a)b^{-1} \subseteq C(bab^{-1}).
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