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Let $Q$ be the quaternion group of order $8$. Write $G = \operatorname{Aut}(Q)$ and $N = \operatorname{Inn}(Q)$. Show that $C_G(N) = N$

What I tried

I know that $C_G(N) = \{g \in G : gN = Ng\}$. I already know that $nN =Nn$ if $n \in N$, which means that $C_G(N) \supseteq N$.

For $C_G(N) \subseteq N$, all I need to show is that $$ \forall g \in G \setminus N, \exists n \in N, \quad gng^{-1} \notin N $$ In other words, I have to show that if $\phi : Q \rightarrow Q$ is an arbitrary isomorfism, that cannot be written as a conjugation, there is some $y \in Q$ such that $$ \text{for each }q \in Q, \qquad \left(x \ \mapsto \ \phi(y\phi^{-1}(x)y^{-1})\right) \quad \neq \quad \left( x \ \mapsto \ qzq^{-1}\right) $$ I rewrote $\phi(y\phi^{-1}(x)y^{-1}) = \phi(y)x\phi(y)^{-1}$

But this is a conjugation right? Did I mess things up somewhere?

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  • $\begingroup$ To show that $C_G(N) \supseteq N$, you have to show that $nm=mn$ for all $n,m \in N$, not just that $nN=Nn$. Also, to show that $C_G(N) \subseteq N$, you have to show that $gng^{-1} \ne n$, not that $gng^{-1} \not\in N$. (In fact, as you have shown, it is true that $gng^{-1} \in N$.) To do this, you will need to work out what $G$ is. $\endgroup$
    – Derek Holt
    Aug 5, 2014 at 20:30
  • $\begingroup$ Other approach - more directly, see also math.stackexchange.com/questions/195932/…, $Aut(Q) \cong S_4$ and $Inn(Q) \cong V_4$. $\endgroup$
    – Nicky Hekster
    Aug 5, 2014 at 20:32
  • $\begingroup$ This one is also relevant: math.stackexchange.com/questions/30108/… $\endgroup$
    – Nicky Hekster
    Aug 5, 2014 at 20:34
  • $\begingroup$ What is $C_G(N)$? Is that the commutator subgroup? $\endgroup$
    – Sidharth Ghoshal
    Dec 20, 2015 at 0:43

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