This is a question from Jacobson's BA 2.
Explanation:
An endomorphism of a group is normal if it commutes with every inner-automorphism of a group $G$ wrt the composition of functions.
In following context, $1_G$ is identity endomorphism. And for any two endomorphism of G, say $f,g$ set $f+g:x\mapsto f(x)g(x)$ (p.s. I don't really like this notation why it uses plus sign even though it is not necessarily commutative)
Anyway, I try to prove:
If $f$ is a surjective normal endomorphism of group $G$, then $f=1_G+g$ where $g$ send $G$ to its center
Update: I have already proved the above proposition. I thought maybe some people do the same question(or homework) in the future and may be confused once as I do, so I leave my answer here rather than delete the post.
if it can be proved that $f$ is idempotent on the derived group $G'$ of group G(i.e. the group generated by the commutator of $G$), then $f\circ f=1_G\circ f $ forall $x$ in $G'$. Due to $f$ is surjective, It would be apparent that $f=1_G$ when $x\in f(G')=G'$(to see $f(G')=G'$, since $f(G')\subseteq G'$, and let $f^{-1}$ be a choice function, then for any $xyx^{-1}y^{-1}=f(f^{-1}(x)f^{-1}(y)(f^{-1}(x))^{-1}(f^{-1}(y))^{-1})$, there element acted by $f$ is apparently in $G'$ so $f(G')\supseteq G'$)
Having that, it became much easier since forall $x,y\in G$ we have $$ xyx^{-1}y^{-1}=f(xyx^{-1}y^{-1})\\ xyx^{-1}y^{-1}f(y)=f(xyx^{-1})=xf(y)x^{-1} \implies x^{-1}y^{-1}f(y)=y^{-1}x^{-1}xf(y)x^{-1}\\\implies x^{-1}(y^{-1}f(y))x=y^{-1}f(y) $$
The fact that $f$ is idempotent on $G'$(or all commutators of $G$) relies on $f$ being normal and being homomorphism: $$ f(xyx^{-1}y^{-1})=f(x)f(y)f(x^{-1})f(y^{})^{-1}=f(x)f(f(y)x^{-1}f(y^{-1}))=f(x)f^2(y)f(x)^{-1}f^2(y^{-1}) \\ =f(f(x)f(y)f(x)^{-1})f^2(y^{-1})=f^2(xyx^{-1}y^{-1}) $$
If $f : G \to G$ is a surjective normal endomorphism, we first need to show that $g^{-1} f(g)$ is central for all $g \in G$, because then $\tilde{f}(g) := g^{-1} f(g)$ defines a map with* $f = \mathrm{id}_G * \tilde{f}$ and $\tilde{f}(G) \subseteq Z(G)$. In fact, something stronger is true: An endomorphism $f : G \to G$ is normal if and only if $g^{-1} f(g)$ centralizes $f(G)$ for all $g \in G$.
The proof of this equivalence is straight forward. $f$ is normal iff $f$ commutes with the inner automorphism $c_g : G \to G$ for each $g \in G$, which means $f(g x g^{-1})=g f(x) g^{-1}$ for all $x \in G$. Since $f$ is a homomorphism, this means $f(g) f(x) f(g)^{-1} = g f(x) g^{-1}$, which can be rewritten as $g^{-1} f(g) \cdot f(x) = f(x) \cdot g^{-1} f(g)$, and we are done.
When $f$ is normal and surjective, the map $\tilde{f}$ is a homomorphism as well: The equation $(gh)^{-1} f(gh) = g^{-1} f(g) h^{-1} f(h)$ simplifies to $h^{-1} g^{-1} f(g)= g^{-1} f(g) h^{-1}$, which holds since $g^{-1} f(g)$ is central in $f(G)=G$.
*I refuse to use that notation $+$ here.