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$Q_8$ is the Quaternion group. In $Q_8$ why $C_G(i)=C_G(-i)$?

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Any particular reason you repeated the question 4 times? Also: what have you thought about so far? Are you familiar with the definition of $Q_8$? Do you know what $C_G$ means? –  Zev Chonoles Feb 25 '13 at 4:58

4 Answers 4

up vote 4 down vote accepted

For any group $G$ and $x \in G$, $C_G(x) = C_G(x^{-1})$.

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In your conclusion, $\forall y\in C_G(x), yx=xy$ means $yx^{-1}=x^{-1}y$. It's right. I got it. –  Ian Feb 25 '13 at 5:18

More generally (if you wish): Show that for any $z\in Z(G)$, the centralizers $C_G(zg)$ and $C_G(g)$ are equal. To do this, argue that $zgc=czg$ if and only if $cg=gc$, using $zc=cz$. This applies in any $G$, and in particular in the quaternion group where $-1$ is central.

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I take it $C_G(x)$ means the centralizer of $x$, which is the elements that commute with $x$. So, which elements commute with $i$? which with $-i$?

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Assume $$Q_8=\langle i,j,k\mid i^2=j^2=k^2=ijk\rangle=\{i,j,k,+1,-1,-i,-j,-k\}$$ in which $$ij=k,ji=-k,~~~~jk=i,kj=-i,~~~~ki=j,jk=-i$$ If you are new to group theory, it's better to do some handy calculations to know this non abelian finite group better. Moreover, you will find out what other answers are trying to tell you via formal theoretical approaches about your question.

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+1 Nice contribution! And good advice, too! –  amWhy Feb 25 '13 at 14:00
    
Mornin Amy, :-) –  B. S. Feb 25 '13 at 14:01

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