How many groups $G_i$ (up to isomorphism) are there such that $G\cong G_i \times \mathbb{Z}$?

Question

Recall that a group $G$ is polycyclic if it has a finite series of subgroups $G = G_0 \rhd G_1 \rhd \cdots \rhd G_k = 1$ for which each factor $G_{i−1}/G_i$ (where $i = 1, \ldots , k$) is finite cyclic or infinite cyclic. A group $G$ is called virtually polycyclic if there exists a polycyclic group $H \lhd G$ such that $G/H$ is finite.

Assume that $G$ is a virtually polycyclic group. How many groups $G_i$ (up to isomorphism) are there such that $G\cong G_i \times \mathbb{Z}$? It is a finite number?

My try: I think it's finite. Since $G\cong G_i \times \mathbb{Z}$ and $G$ is virtually polycyclic, $G_i \times \mathbb{Z}$ is virtually polycyclic. So we can comclude that $G_i$'s are virtually polycyclic with Hirsch length $<h(G)$. So the class of $G_i$'s is classified into finitely many subclasses $\mathcal{W}_i$ in each of which two groups have the same Hirsch length. My probelm is that I can't prove that every class has a finite number of elements.

Answer 1

Here is another attempt at an answer! We claim now there are only finitely many possible isomorphism types for the groups $A_c$ in direct decompositions $G = A \times A_c$ with $A \cong ({\mathbb Z},+)$.

Let's call an infinite cyclic direct factor $A$ of $G$ a $Z$-factor and a corresponding factor $A_c$ a complementary factor. Note that all $Z$-factors lie in $Z(G)$ and the commutator subgroup $[G,G]$ is contained in all complementary factors.

Let $N/[G,G]$ be the torsion subgroup of $G/[G,G]$. Then it is easy to see that $N$ is contained in all complementary factors, and the quotient group $G/N$ is finitely generated free abelian and hence isomorphic to ${\mathbb Z}^r$ for some $r$, and every $Z$-factor intersects $N$ trivially.

Let $X$ be the subgroup of $G$ generated by all $Z$-factors. So $X \le Z(G)$. As a subgroup of the free abelian group $G/N$, we can choose a free basis $\bar{x}_1,\bar{x}_2,\ldots,\bar{x}_r$ of $G/N$ such that $XN/N$ is generated by $d_1\bar{x}_1,\ldots,d_s\bar{x}_s$ for some $s$ with $1 \le s \le r$, where $d_i$ are positive integers with $d_i|d_{i+1}$ for all $i$.

For a $Z$-factor $A$, the quotient group $AN/N$ is a subgroup of $\langle \bar{x}_1,\ldots,\bar{x}_s \rangle$, and so it has a complement in $\langle \bar{x}_1,\ldots,\bar{x}_s \rangle$ that is generated by $s-1$ elements $\bar{y}_1,\ldots,\bar{y}_{s-1}$ of $\langle \bar{x}_1,\ldots,\bar{x}_s \rangle$, and then $\langle \bar{y}_1,\ldots,\bar{y}_{s-1},\bar{x}_{r+1},\ldots,\bar{x}_r \rangle$ is a complement of $AN/N$ in $G/N$. Inverse images of these elements of these generators together with $N$ generate a complementary factor $A_c$ for $A$ in $G$ (i.e. $G = A \times A_c$).

We claim now that the isomorphism type of $A_c$ depends only on the values of the elements $\bar{y}_i$ modulo $XN/N$ and, since these values lie in a finite abelian group of order $d_1d_2\cdots d_s$, there are only finitely many possible isomorphism types for $A_c$.

To prove the claim, consider a presentation of $A_c$ over a generating set that includes inverse images $y_1,\ldots,y_s$ of the $\bar{y}_i$. Since the $y_i$ have infinite order, for each $i$, the exponent sum of $y_i$ in any defining relator in the presentation must be $0$. If we replace some $y_i$ by another generator with the same image modulo $XN$, then we are just multiplying the original $y_i$ by an element of $XN$. Multiplying $y_i$ by an element of $N$ does not change $A_c$ (which contains $N$), so we get all other complements of this type by multiplying $y_i$ by elements of $X$. But since $X \le Z(G)$, the defining relators in the original presentation continue to hold, so the new complement has the same presentation as $A_c$ and hence is isomorphic to $A_c$.