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For any group $G$, consider the map $f: G \to \text{Aut($G$)}$ with $g \mapsto f_g (h) = ghg^{-1}$. . This map represents a homomorphism from $G$ into itself, but that it satisfies the requirements are not immediately clear to me.

For example, it must be the case that $f$ maps the identity in $G$ to the identity in $\text{Aut($G$)}$. The identity in $\text{Aut($G$)}$ is the trivial map $f_e (h) = ehe^{-1} = h$. Since identities are unique, this must be the unique identity. However, the kernel of $f$ is the center of $G$, since if $g$ commutes with everything in the group, $ghg^{-1} = h$ for all $h \in G$.

I'm having trouble grasping this. The identity map in $\text{Aut($G$)}$ should not depend on the particular element of $G$, but by selecting $g$ that commute, it seems that we can induce an identity map.

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    $\begingroup$ for all $h\in G$, $f_e(h)=ehe^{-1}=\color{red}h$; the point is that $f_e$ is the identity map; you are correct that other elements $z\in G$ could be mapped to $f_z(h)=f_e(h)$, but that's not a problem -- there could be a non-trivial kernel -- more than one element of $G$ could get mapped to the identity element in Aut($G$), but that does not mean there is more than one identity element in Aut($G$) $\endgroup$ Feb 23 '20 at 3:54
  • $\begingroup$ Fixed, thank you. $\endgroup$
    – John P.
    Feb 23 '20 at 3:55
  • $\begingroup$ Just to be sure I understand: say that $f$ has a non-trivial kernel consisting of $g$ and $e$. So $g \mapsto f_g(h) = h, \; \forall h$. Then, because the identity in $\text{Aut($G$)}$ is unique, it must be the case that $f_g = f_e$, even if $g \neq e$? $\endgroup$
    – John P.
    Feb 23 '20 at 4:06
  • $\begingroup$ Although, if that's the case, wouldn't f not be injective? I suppose we never required that to be the case, so this is actually fine. $\endgroup$
    – John P.
    Feb 23 '20 at 4:07
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    $\begingroup$ isomorphisms must be injective, but not homomorphisms; in fact, if $G$ is Abelian, then $f$ maps $g$ to the identity in Aut($G$) for all $g\in G$ $\endgroup$ Feb 23 '20 at 4:12
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Firstly, I think that a writing like "$g \mapsto f_g (h) = ghg^{-1}$" is misleading: $f$ takes an element of $G$ to an automorphism of $G$, not to another element of $G$. So, a better way is: $g \mapsto (h \mapsto f_g (h) := ghg^{-1})$. (Or, even better, use stacked maps.)

Said this, it's a fact that:

$$g \in Z(G) \Rightarrow f_g(h)=ghg^{-1}=hgg^{-1}=h, \color{red}{\forall h \in G} \Rightarrow f_g=\iota_G$$

IMO, you should restart from this to see what doesn't still convince you (if any).

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$f$ maps $e$ to the identity automorphism, but it also maps any other element of the center to the identity.

For instance, the Klein four group is abelian, so $f$ would map everything to the identity (not just $e$, as it must to be a homomorphism).

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