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I will take as definition outer automorphisms as any automorphism that is not inner. We say that a group $G$ is complete if it is centerless and $G$ admits no outer automorphisms.

As a fact, $G$ being complete implies that $G\cong Aut(G)$. I’ve been wondering whether the converse is true. If we do not assume $G$ is centerless, then a quick counterexample is $D_8$. So my question is, if $G$ is centerless, does $Aut(G)\cong G$ imply that $G$ is complete?

By a size argument, this is true if $G$ is finite, but I haven’t made progress on other cases. Any hint is appreciated.

Thomas Andrews
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Meow
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  • Kenta S has provided a counterexample to the statement as posed. However, a modification of the statement is true: $G$ is complete if the natural map $G \to Aut(G)$ is an isomorphism. (Note that Kenta's example has $G\cong Inn(G) \subset Aut(G)$ as a proper subgroup while nevertheless $G \cong Aut(G)$) – Noah Caplinger Dec 22 '24 at 15:00

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Let $D=\langle s,t|s^2=t^2=1\rangle=C_2*C_2$ be the infinite dihedral group. Then this answer shows that $$\mathrm{Aut}(D)=D\rtimes C_2$$ where $C_2$ exchanges the two factors of $C_2$. Explicitly, $$D\rtimes C_2=\langle s,t,i:s^2=t^2=i^2=1,isi=t\rangle.$$ But there is an isomorphism $$D\simeq D\rtimes C_2$$ given by $s\mapsto s$ and $t\mapsto i$. The inverse is given by $s\mapsto s$, $t\mapsto isi$, and $i\mapsto t$.

Intuitively, there is an isomorphism because in the relations, where $t=isi$, the relation $t^2=1$ is superfluous so $t$ can just be ignored: $$\langle s,t,i:s^2=t^2=i^2=1,isi=t\rangle=\langle s,i:s^2=i^2=1\rangle.$$

Kenta S
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