I know $Z(\operatorname{Aut}(G))=\{e\}$ , then $Z(G)$ may not be identity. As, for example, take $G = V$ (Klein's $4$-group). Then $\operatorname{Aut}(G)=S_3$. Hence $Z(\operatorname{Aut}(G))=\{e\}$ but $Z(G) =G$. (Here $Z(G)$ means center of $G$) My question is the converse of the result. I know this is true but How can I proceed? Any hint?
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2How do you know this is true? Why can't you write down the argument that convinces you? – Derek Elkins left SE May 31 '19 at 05:50
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If $\phi$ is our central automorphism, then since it commutes with the conjugation automorphisms, we have for all $g,h\in G$, $$\phi(ghg^{-1})=g\phi(h)g^{-1}$$
Since $\phi$ is an automorphism, we can expand and rearrange to obtain, for all $g,h\in G$:$$\big(g^{-1}\phi(g)\big)\phi(h)\big(g^{-1}\phi(g)\big)^{-1}=\phi(h)$$
So for any fixed $g$, the element $g^{-1}\phi(g)$ commutes with every element in $\phi(G)=G$, so since $Z(G)={e}$, we have that $g^{-1}\phi(g)=e$ for all $g\in G$.
Thus, our central automorphism $\phi$ must be the identity.
Chris H
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Then every central automorphism is identity. Then how can u say all automorphism belonging to centralizer is identity? – Pradip May 31 '19 at 10:27
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1A central automorphism is just what I am calling an element of the centre of $Aut(G)$, so we can deduce $Z(Aut(G))=e$. – Chris H May 31 '19 at 11:37
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