# Confusion on homomorphism map

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.

• 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($G$) Feb 23 '20 at 3:54
• Fixed, thank you. Feb 23 '20 at 3:55
• 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 $g \neq e$? Feb 23 '20 at 4:06
• 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. Feb 23 '20 at 4:07
• 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 $g\in G$ Feb 23 '20 at 4:12

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