Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 3 / 6D / Solution

For fixed $g\in G$, conjugation $g_{\ast }:x\mapsto gx{g}^{-1}$ preserves multiplication:and its inverse is $(g^{-1}{)}_{\ast }$. It is therefore a group automorphism. Furthermore,so $\theta :g\mapsto g_{\ast }$ is a homomorphism $G\to \mathrm{Aut}(G)$. Its kernel consists of elements commuting with every $x$, hence

For $\alpha \in \mathrm{Aut}(G)$,so the inner automorphism group is normal in the automorphism group.

If $G=⟨x⟩$ is cyclic, every automorphism has the form $x\mapsto x^a$. Two such maps commute because $ab=ba$, so $\mathrm{Aut}(G)$ is abelian. The conclusion fails for general abelian groups: for $G=C_2\times C_2$, its three nonidentity elements may be permuted arbitrarily, givingwhich is nonabelian.