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I am watching Benedict Gross's abstract algebra lectures, and wanted to clarify with someone one comment he makes.

He defines the mapping $f: G \to \text{Aut $G$}$ for some group $G$ and proves that it is a homomorphism. He establishes, generally, that its kernel is the center of $G$. All of this makes sense to me. But, instead of defining the image of $f$ generally, he gives an example of the Klein four-group, whose image contains only the identity element.

My question is: is there a way to define the image of this mapping generally for any $G$? Does it depend on $G$, even though the kernel does not?

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    $\begingroup$ The image is called the group of inner automorphisms of $G$ (assuming the map sends the element $g$ to the automorphism of conjugation-by-$g$, $g\mapsto \varphi_g$ with $\varphi_g(x) = gxg^{-1}$). It is of course isomorphic to $G/Z(G)$, so it will depend on what $G$ is. It may or may not be all of $\mathrm{Aut}(G)$. $\endgroup$ – Arturo Magidin Feb 22 '20 at 4:56
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As @Arturo points out the image is the group of so-called inner automorphisms, which are defined by conjugation by elements of $G$. For some groups, like all symmetric groups except $S_2$ and $S_6$, it's the entire automorphism group.

The Klein four group is abelian, which explains the example you mentioned: the center is everything.

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