# $Aut(G)$ is abelian if and only if $G$ is cyclic. [duplicate]

Problem says:

Let G be an abelian group. Prove that Aut(G) is abelian if and only if G is cyclic.

And I solved $(\Leftarrow)$ direction as follow:

Suppose that $G$ is cyclic. Then $Aut(G)=\{\varphi_{1},\cdots,\varphi_{n}\}$ where $\varphi_{i}(g)=g^{i-1}$ . Thus, for $\varphi_{i},\varphi_{j}\in Aut(G)$ , $\varphi_{i}\varphi_{j}(g) = \varphi_{i}(g^{j-1}) = (g^{j-1})^{i-1} = (g^{i-1})^{j-1} = \varphi_{j}(g^{i-1}) = \varphi_{j}\varphi_{i}(g)$

Therefore, $\varphi_{i}\varphi_{j}=\varphi_{j}\varphi_{i}$ for $1\leq i,j\leq n$ . Hence, $Aut(G)$ is abelian.

But I worried about the case where $G$ is infinite and the other direction.

The first implication is also true, i.e., a finite abelian group with abelian automorphism groups is cyclic (see this duplicate MSE-question). In general however, a group with abelian automorphism group need not be abelian, i.e., in particular not cyclic. See the article Jonah, D.; Konvisser, M. Some non-abelian $p$-groups with abelian automorphism groups. Arch. Math. (Basel) 26 (1975), 131--133.