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

Question

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.

Answer 1

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.

In the infinite case, there are locally cyclic groups that are not cyclic, and these have abelian automorphism groups. For instance, the additive group of rational numbers has an abelian automorphism group (the multiplicative group of rational numbers).