Given a Group $G$, the $n$-th Engel Word defined by $[x,\ _0{x}]=x: \ [x,_{ \ n}y]=[[x,_{\ {n-1}}y],y]$ in the group $G$ consists by substituting group elements for the determinates. The Group generated by that words is called the $n$-Engel verbal subgroup of $G$ and is denoted by $E_n(G)$. So there exists a normal series of subgroups $$G=E_0(G) \geq E_1(G) \geq E_2(G) \geq \ldots \geq E_n(G) \geq \ldots \geq 1$$
A group $Q$ is called Chernikov group if it has a subgroup of finite index that is a direct product of finitely many groups of type $C_{p^{\infty}}$ for various prime $p$ (quasicyclic $p$-groups).
Assume that the group $E_k(G)$ is not be nilpotent, soluble or $m$-Engel for some $m$ (this is important, cause in my hipotesis in the others results this is not possible)
This is my question: Let $G$ be a group, if the term $E_k(G)$ of $n$-Engel series of the group $G$ is generated of Chernikov groups and if the term of the series will be stabilized, i.e, $E_k(G)=E_{k+1}(G)$. Can I to prove that $E_k(G)=(E_k(G))'$?
My ideias: I have worked try to proof assuming that $E_k(G)$ is not be perfect, then a group $E_k(G)$ contain a maximal perfect subgroup that is totally invariant. If $E_k(G)$ is not soluble, then $E_k(G)$ is not abelian. Suppose that $E_k(G)$ is simple, then $R=E_k(G)$ and so $E_k(G)$ is perfect. So suppose that $E_k(G)$ is not simple, I thought to take a normal subgroup $N$ of $E_k(G)$, Can I suppose that $N$ is Chernikov group? There are other forms to resolve?