1
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
0

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.