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Let $G$ be finitely generated group, suppose that $G$ has a residually nilpotent subgroup $N$ of finite index in $G$ , we wont to show that $N$ contains characteristic subgroup of finite index in $G$.

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Residual nilpotency has nothing to do with this. Every finite index subgroup $N$ of every finitely generated group $G$ contains a characteristic subgroup of finite index. To see why, each automorphism $\phi : G \to G$ takes $N$ to a subgroup $\phi(N)$ of the same index, $[G:N] = [G:\phi(N)]$. Second, the number of subgroups of a fixed finite index is finite (this is where you use that $G$ is finitely generated). Third, the subgroup $$\Gamma = \cap_{\phi \in \text{Aut}(G)} \, \, \phi(N) $$ is characteristic and is of finite index (because it is an intersection of finitely many finite index subgroups), and it is contained in $N$.

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