I'm reading an article I wrote my doctoral supervisor. In this article he states that if $G$ be a hypercentral group and suppose that $G$ is generated by (a finite number of) Prufer subgroups. Then $G$ is abelian.


I asked him about this statement. He told me it was true. He said that in the good books of reference I should find a demonstration of this fact. Turns out I do not find this fact anywhere. Can anyone give me some information about this statement? Or any idea of the proof of this statement.
Thank you so much.



A condition for Hypercentral Groups to be Abelian
This should follow along the lines of Robinson's book "Finiteness conditions and generalized soluble groups", Part 2, Section 9.2. I will try to sketch an argument using Robinson's terminology. Since $G$ is hypercentral, it is locally nilpotent. By your assumption it is therefore periodic. Since it is generated by Pruefer groups, it is semi-radicable, that is, $G=G^n$ for every $n$. In hypercentral groups, being semi-radicable and radicable is the same (Robinson, Corollary 1 on p. 125), and Corollary 2 therein implies that $G$ is abelian.


