A group $G$ is said to be radicable if each element is an $n$th power for every positive integer $n$, ie, $G$ is radicable if the equation $x^n = a$ has a solution in $G$ for every positive integer $n$ and every $a \in G$.

I'm reading the paper "A finiteness condition for verbal conjugacy classes in a group" by J.M Muñoz-Escolano and P. Shumyatsky and in this paper, in the Lemma 2.6 they consider a group $A$ radicable and Chernikov and claim that such a group has no subgroup of finite index.

My question is: Radicable Groups has no subgroup of finite index? Why? Or the hypothesis that the group is required to be Chernikov for this to be valid? In any case as proof this statement?