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Let $G$ be a nilpotent group. That is, the lower central series $G,[G,G],[[G,G],G],\dotsc$ terminates.

Denote the $n$-th term of this series by $\gamma_n(G)$ (in particular, $\gamma_1(G)=G$).

It is clear that every $\gamma_n(G)$ is normal in $G$ (since $\gamma_n(G)$ is generated by $n$-fold commutators, and conjugation preserves $n$-fold commutators set-wise).

Now, assume further that each quotient $\gamma_n(G)/\gamma_{n+1}(G)$ is a finitely generated free abelian group (that is, isomorphic to $\mathbb{Z}^{d_n}$ for some $d_n$).

Then, we can refine the lower central series to a series with infinite cyclic quotients.

My question is whether we can arrange a refinement such that every subgroup in the refinement is normal in $G$.

It seems to me that the answer is yes, and in fact, this will be true no matter how we refine the lower central series to a series with infinite cyclic quotients.

However, since I have never seen this claim, I would like to make sure it's true.

My reasoning is that for every $n$, every subgroup $H$ of $G$ between $\gamma_{n+1}(G)$ and $\gamma_n(G)$ is normal in $G$ because conjugation of an element $h$ of $H$ by an element of $G$ is equivalent to multiplication of $h$ by some element of $\gamma_{n+1}(G)$.

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