Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
up vote 4 down vote accepted

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

share|cite|improve this answer
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.

share|cite|improve this answer

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 $-i$?

share|cite|improve this answer

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.

share|cite|improve this answer
+1 Nice contribution! And good advice, too! – amWhy Feb 25 '13 at 14:00
Mornin Amy, :-) – Babak S. Feb 25 '13 at 14:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.