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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- Oct 20th 2008, 05:29 PMGreenandGoldAbstract 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!!! :)

- Oct 22nd 2008, 01:22 AMPrometheus
I didn't quite understand what you meant, but you should solve this using the definition of centralizers

$\displaystyle C(b^{-1}ab)=\{g \in G | gb^{-1}ab = b^{-1}abg \} = \{g \in G | bgb^{-1}abb^{-1} = bb^{-1}abgb^{-1} \} $$\displaystyle = \{g \in G | bgb^{-1}a = abgb^{-1} \} = \{g \in G | bgb^{-1} \in C(a) \} $$\displaystyle = \{g \in G | g \in b^{-1}C(a)b \} = b^{-1}C(a)b $

so you get $\displaystyle C(b^{-1}ab) = b^{-1}C(a)b $ - Oct 22nd 2008, 05:21 PMThePerfectHacker