# Let $G$ be a polycyclic group and assume that every finite quotient of $G$ is nilpotent. Then $G$ is nilpotent

Let $$G$$ be a polycyclic group and assume that every finite quotient of $$G$$ is nilpotent.

Then $$G$$ is nilpotent.

First some preliminaries:

1. Every infinite polycyclic group contains a free abelian normal subgroup.

2. On a polycyclic group one can use 'Noetherian' induction:

If $$G$$ is polycyclic and doesn't have a property P, we can assume every proper quotient of $$G$$ has the property P, while $$G$$ doesn't. (see the book Polycyclic groups by Daniel Segal)

1. If $$G$$ is finitely generated and nilpotent of class $$k$$, we can define the Hirsch length of $$G$$ as the sum $$\sum_{i =1}^{k}m_i = h(G)$$ where $$m_i$$ is the rank of the free part of $$C^i(G)/C^{i+1}(G)$$.

2. If $$G$$ is polycyclic, then we can define the Hirsch number as the number of infinite factors in a witness series of $$G$$'s polycyclicity. This number is the same for all such series, and it equals the Hirsch length from above when $$G$$ is also nilpotent (and hence finitely generated).

Here's where I got to:

If $$G$$ is finite, by assumption it is nilpotent. So we can assume $$G$$ is infinite.

Let $$A \triangleleft G$$ be a free abelian subgroup.

Let $$p$$ denote a prime number; if $$G/A^p$$ is trivial, $$G$$ is abelian and hence nilpotent.

So we can assume that for every $$p$$, $$G/A^p$$ has the property that if every finite quotient is nilpotent, then so is $$G/A^p$$.

But if $$T/A^p \triangleleft G/A^p$$, and $$G/A^p / T/A^p \cong G/T$$ is a finite quotient, by assumption $$G/T$$ is nilpotent. Therefore for every $$p$$, $$G/A^p$$ is nilpotent.

Thus, $$\forall p$$ prime $$\exists n_p \in \mathbb{N}$$ with $$C^{n_p}(G) \subset A^p$$.

Now, each such $$G/A^p$$ is polycyclic, and nilpotent, and hence finitely generated. Thus $$h(G/A^p) \leq h(G)$$. From here I suspect we can bound the sequence $$n_p$$ to get that there is $$m \in \mathbb{N}$$ s.t $$C^{m}(G) \subset \cap_{p} A^p$$.

Questions:

1. Can I bound this sequence $$n_p$$?

2. Why is it true that $$\cap_p A^p = \{e\}$$?

I guess $$C^i(G)$$ is the lower central series of $$G$$?
Since $$G/A$$ is nilpotent, we have $$C^m(G) \le A$$ for some $$m$$. Now, for any prime $$p$$, $$A/A^p$$ is elementary abelian of order $$p^d$$, where $$A$$ is free abelian of rank $$d$$. So the maximum length of a chain of subgroups of $$A/A^p$$ is $$d$$. Hence, since $$G/A^p$$ is nilpotent, we must have $$C^{m+d}(G) \le A^p$$. This answers your first question.
Your second question is easy, because elements of $$A$$ have the form $$(x_1,\ldots,x_d)$$ for some $$x_i \in {\mathbb Z}$$, and if $$(x_1,\ldots,x_d) \in \cap_{p}A^p$$, then $$p \vert x_i$$ for each $$i$$ and all primes $$p$$, so $$x_i=0$$.