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let $N$ be nilpotent subgroup of finite index in finitely generated abelien by nilpotent(i.e there exist normal subgroup $M$ abelien such that $G/M$ is nilpotent ), proceed by induction on the order of $G/N$ and consider first of all the special case where $G/N$ cyclic i wont to prove that there exist normal subgroup $H$ of index some prime $q$ in $G$ where $H$ contain $N$.

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There's a bit of redundancy here, it seems.

Assume $G$ is abelian-by-nilpotent, so that there is an abelian normal subgroup $M$ such that $G/M$ is nilpotent. Let $N$ be a proper, normal subgroup of finite index.

Then $G/N \ne 1$ is also abelian-by-nilpotent. This is because

  • $$M N / N \cong M /(M \cap N)$$ is a normal subgroup of $G/N$ which is abelian, as a quotient of the abelian group $M$, and
  • $$ (G / N) / ( M N / N) \cong G /M N \cong (G/M) / (MN /M) $$ is nilpotent, as a quotient of the nilpotent group $G / M$.

Now all you have to prove is that a finite, abelian-by-nilpotent group $H \ne 1$ has a normal subgroup of index a prime. To do this, just note that $H' < H$.

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  • $\begingroup$ thanks, please i wont to ask why $H' < H$ gives $H≠1$ has a normal subgroup of index a prime $\endgroup$ – user220373 Mar 21 '15 at 11:24
  • $\begingroup$ Becase $H/H'$ is now a nontrivial finite abelian group, and thus it has definitely a subgroup of index a prime. $\endgroup$ – Andreas Caranti Mar 21 '15 at 11:26
  • $\begingroup$ @user220373, you're welcome. $\endgroup$ – Andreas Caranti Mar 21 '15 at 11:54

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