5
$\begingroup$

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

$\endgroup$
  • 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 2-step 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
7
$\begingroup$

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 torsion-free 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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.