Normal automorphism of a perfect group Show that the only normal automorphism (i.e., commutes with every inner automorphism) of a perfect group is the identity automorphism. 
 A: Given $g \in G$ let $c_g: G \to G$ be the conjugation by $g$, i.e., $c_g(h) = ghg^{-1}$ for $h \in G$. Then, $c_g = c_{\phi(g)}$ since $\phi c_g \phi^{-1} = c_{\phi(g)}$ and $\phi$ commutes with $c_g$.
Therefore, $ghg^{-1} = \phi(g)h \phi(g)^{-1}$ for all $g, h \in G$. We get $g^{-1}\phi(g) \in Z(G)$. Note that $g \mapsto g^{-1}\phi(g)$ is a homomorphism from $G$ to $Z(G)$, which is abelian. Therefore it factors through $G^{ab} = 1$.
We conclude that $g = \phi(g)$ for all $g$.
A: If f is a normal automorphism on G then f(a^-1ba)=a^-1f(b)a for all a,b in G. Let c(a)=f(a)a^-1. Then c(a) is in Z(G), the center of G, since if g is in G, there is an h in G where g=f(h). c(a)g = f(a)a^-1f(h)aa^-1 = f(a)f(a^-1)f(h)f(a)a^-1 = f(h)f(a)a^-1 = gc(a). G is perfect means every member of G is a product of commutators, each of the form aba^-1b^-1. If f is the identity map on every commutator then it is the indentity map on G. f(aba^-1b^-1) = f(a)f(ba^-1b^-1) = f(a)bf(a^-1)b^-1 = ac(a)b((c(a)^-1)a^-1b^-1 = aba^-1b^-1.
