$\rm{Aut}(G)$ is not cyclic when $G$ is not abelian

Question

we have a $G$ which is not abelian and we need to prove that the group of all automorphisms is not cyclic.

any ideas?

Try this: If $\rm{Aut}(G)$ is cyclic, then $G$ is an abelian group and If $G/Z(G)$ is cyclic then $G$ is abelian. — Bobby, Jan 27 '14 at 13:01

score 3 · Answer 1 · answered Jan 27 '14 at 13:02

3

Hint: If $G/Z(G)$ is cyclic then $G$ is abelian.

answered Jan 27 '14 at 13:02

we tried using it. but we don't get why the automorpism group is isomorphic to G/Z(G) – idan Jan 27 '14 at 13:10
@idan: $$G/Z(G)\cong Inn(G)\le Aut(G)$$with $;Inn(G);$ being the group of inner automorphisms of $;G;$ . – DonAntonio Jan 27 '14 at 13:19
got it! thanks! – idan Jan 27 '14 at 13:24

1 Answers1