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

group-theoryinfinite-groupsnilpotent-groups

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 $\cap_p A^p = \{e\}$?

Best Answer

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 $x_i=0$.

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