TITLE: Normal automorphism of a perfect group
QUESTION [2 upvotes]: Show that the only normal automorphism (i.e., commutes with every inner automorphism) of a perfect group is the identity automorphism.

REPLY [0 votes]: 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.
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