# $C_{\operatorname{Aut}(Q)}(\operatorname{Inn}(Q)) = \operatorname{Inn}(Q)$?

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?

• 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. Aug 5, 2014 at 20:30
• Other approach - more directly, see also math.stackexchange.com/questions/195932/…, $Aut(Q) \cong S_4$ and $Inn(Q) \cong V_4$. Aug 5, 2014 at 20:32
• This one is also relevant: math.stackexchange.com/questions/30108/… Aug 5, 2014 at 20:34
• What is $C_G(N)$? Is that the commutator subgroup? Dec 20, 2015 at 0:43