Let $G$ be a abelian and radicable group generated by subgroups abelian, radicable and Chernikov. Then $G$ is Chernikov? ie, if $G = \left<H| H \leq G \right>$ with $H$ abelian, radicable and Chernikov, then $G$ is Chernikov?

Remark: 1) This seems to me true since the number of subgroups $H$ is finite. But I do not know if it is true in general;

2) A group $G$ is radicable (by D. Robinson, for example) if each element is an $n$-th power for every positive integer $n$. In particular, radicable groups has no subgroup non-trivial of finite index;

3) A group which is a finite extension of an abelian group satisfying Min is called a Chernikov group. In other words, $G$ is a Chernikov group, if has a subgroup $N$ of finite index such that $N$ is the direct product of a finite number of quasicyclic $p$-groups ($C_{p^\infty}$ ).


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