# Abstract Algebra

• Oct 20th 2008, 05:29 PM
GreenandGold
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!!! :)
• Oct 22nd 2008, 01:22 AM
Prometheus
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 $C(b^{-1}ab) = b^{-1}C(a)b$
• Oct 22nd 2008, 05:21 PM
ThePerfectHacker
Quote:

Originally Posted by GreenandGold
A little help needed. Prove: C(bab^-1)=bC(a)b^-1.

Just show $C(bab^{-1})\subseteq bC(a)b^{-1}$ and $bC(a)b^{-1} \subseteq C(bab^{-1})$.