# Image of homomorphism $f: G \to \text{Aut$G$}$

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?

• 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 $\mathrm{Aut}(G)$. – Arturo Magidin Feb 22 '20 at 4:56

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.