Is there any infinite noncyclic group whose automorphism group is abelian..can we find a sufficient condition for infinite group to have an abelian automorphism group Thank you

4$\begingroup$ Group of integers? $\endgroup$ – Mohan Jun 9 '19 at 17:43

1$\begingroup$ @Mare obviously not. A group with abelian automorphism group has to be 2step nilpotent. $\endgroup$ – YCor Jun 9 '19 at 17:48

$\begingroup$ Every cyclic group has abelian automorphism group..i am seeking for a non cyclic infinite group or a sufficient condition that makes an infinite group to be a Miller group $\endgroup$ – Mohammad Radi Jun 9 '19 at 17:49

4$\begingroup$ Additive group of rationals has abelian automorphism group too, and is not cyclic. $\endgroup$ – Wojowu Jun 9 '19 at 18:06

1$\begingroup$ @YCor So presumably you consider abelian groups to be $2$step nilpotent? (Perhaps $n$step nilpotent does not mean the same as nilpotent of class $n$.) $\endgroup$ – Derek Holt Jun 9 '19 at 19:06
The finite abelian groups that can be the automorphism group of an infinite abelian group have been classified by Fournelle in [Finite groups of automorphisms of infinite groups II, J. of Algebra 80, 1983, 106  112, Theorem 1.2]:
There is an infinite abelian group $A$ with $Aut(A) = G$ for a finite abelian group $G$ iff $G$ is of even order and is a direct product of cyclic groups of orders 2, 3, and 4 with the property that if $G$ has an element of order 12 it also has an element of order 2 that is not a sixth power.
Examples of torsionfree groups $A$ with $Aut(A)=G$ for $G$ as above are constructed in [Fuchs: Infinite abelian groups II, Chap. XVI, Sect. 116] in examples 1, 2 and Theorem 116.2.