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In the finite group presentation $⟨a,t\mid ta{t}^{-1}=a^2⟩$, let $a$ act by $x\mapsto x+1$ and $t$ by $x\mapsto 2x$. The conjugates $a_i=t^{-i}a{t}^i$ satisfy $a_{i+1}^2=a_i$, so their increasing cyclic subgroups give the additive group $\mathbb{Z}[1/2]$. Every element has the form $a_i^m{t}^k$, and the affine action proves that this form can be the identity only if $k=m=0$. Thus the displayed semidirect product identification holds. This is a finitely presented group with a series having torsion-free abelian factors, but it is not a poly-infinite cyclic group, since $\mathbb{Z}[1/2]$ is not finitely generated.

The Hirsch length of a polycyclic group counts infinite cyclic factors in a polycyclic series. Refinement shows it is independent of the series, additive in extensions, and unchanged in finite-index subgroups. A finite-index poly-infinite cyclic group has one infinite cyclic factor for each unit of Hirsch length.

A nilpotent group has lower central series ${\gamma }_{1}G=G$, ${\gamma }_{j+1}G=[G,{\gamma }_{j}G]$ terminating at the identity. A polycyclic group has a finite subnormal series with cyclic factors, which may be finite or infinite.

If a nilpotent group is generated by finitely many elements, then ${\gamma }_{j}G/{\gamma }_{j+1}G$ is generated by the finitely many weight-$j$ iterated commutators in these generators. Modulo the next term, commutator identities make these quotients abelian and permit this collection of words. Each quotient is thus a finitely generated abelian group. Refine each by a series with cyclic factors and splice the refinements through the finite lower central series. This proves that a finitely generated nilpotent group is polycyclic.

For the duality-group assertion, first prove the useful finite-index reduction. Induct on the length of a polycyclic series, taking its last proper term $H◃G$. By induction $H$ has a finite-index poly-infinite cyclic group. Because $H$ is finitely generated, it has only finitely many subgroups of any fixed finite index. Intersect all automorphic images of that subgroup to obtain a characteristic finite-index subgroup $H_0\subset H$ which is still poly-infinite cyclic. It is normal in $G$. The quotient $G/H_0$ is finite-by-cyclic. If it is finite, $H_0$ already suffices; if its cyclic quotient is infinite, the subgroup generated by a lift of its generator is infinite cyclic and has finite index. Its preimage in $G$ is an extension of $H_0$ by $\mathbb{Z}$, hence poly-infinite cyclic. This proves polycyclic groups are virtually poly-infinite cyclic.

Take such a finite-index subgroup $P$ of the given dimension-three Poincare duality group. It is again a dimension-three duality group. The extension rule proved in Question 3, applied successively to its infinite cyclic series, makes $P$ a duality group of dimension equal to its Hirsch length. Hence that length is three. Its last quotient giveswhere $H$ has a poly-infinite cyclic series of length two. Thus $H\cong \mathbb{Z}⋊\mathbb{Z}$, with the last generator acting on the first by $+1$ or $-1$. In the first case $H={\mathbb{Z}}^2$. In the second case $H$ is the Klein-bottle group $⟨a,b\mid ba{b}^{-1}=a^{-1}⟩$. Its characteristic subgroup $⟨a,b^2⟩\cong {\mathbb{Z}}^2$ is the centralizer of its commutator subgroup $⟨a^2⟩$. Replace $H$ by this characteristic index-two subgroup if necessary. It remains normal in $P$, and the quotient is finite-by-infinite-cyclic. Taking the preimage of a finite-index infinite cyclic subgroup givesThis extension splits: any lift $t$ of the cyclic generator supplies a section. Therefore $P_0={\mathbb{Z}}^2{⋊}_B\mathbb{Z}$ for some $B\in G{L}_2(\mathbb{Z})$.

In additive notation on ${\mathbb{Z}}^2$, $[t,v]=(B-I)v$. All its other internal commutators vanish, so induction givesConsequentlyThis is the nilpotence criterion for an abelian-by-cyclic group. If $B=I$ the group is abelian; otherwise it is two-step nilpotent. The criterion concerns this actual semidirect product. A finite extension of it is only known to be virtually nilpotent: for example ${\mathbb{Z}}^2{⋊}_{-I}\mathbb{Z}$ is virtually abelian but its lower central series never terminates.

Write elements of the real Heisenberg group as $(x,y,z)$. Direct matrix multiplication and inversion giveThese formulas show that $N_k=(k\mathbb{Z}{)}^3$ is a subgroup. It is discrete. For the right quotient, multiplying by $(ka,kb,kc)$ changes the coordinates toThus $(x\mathrm{mod}k,y\mathrm{mod}k)$ is well defined on the quotient. Over a small base rectangle choose real lifts $x,y$ of its coordinates. Every coset above that rectangle has a representative $(x,y,z)$, and two such representatives differ precisely by $z\mapsto z+kc$. Henceis a smooth local trivialization. The fibres are circles, proving the locally trivial fibration over ${\mathbb{R}}^2/k{\mathbb{Z}}^2\cong T^2$. Reducing first $x,y$ and then $z$ also gives a compact fundamental region, so $N_k$ is cocompact.

The ambient Lie group is diffeomorphic to ${\mathbb{R}}^3$, hence simply connected and contractible. Its quotient has fundamental group $N_k$, and the first integral homology is its abelianization. Let $a=(k,0,0)$, $b=(0,k,0)$ and $c=(0,0,k)$. Every element is uniquely $a^m{b}^n{c}^{\ell }$, since its third coordinate is $k^{2}mn+k\ell$. The commutator formula isso $c$ is central and $[a,b]=c^k$. Every commutator lies in $(0,0,k^2\mathbb{Z})$, and this subgroup is generated by $[a,b]$. Thus $N_k^{\prime }=⟨c^k⟩$. The homomorphism $(x,y,z)\mapsto (x/k,y/k,z/k\mathrm{mod}k)$ is surjective onto ${\mathbb{Z}}^2\oplus \mathbb{Z}/k$ and has exactly this kernel; the cross term in multiplication is divisible by $k$ after dividing by $k$. ThereforeFor $k=1$ the last factor is trivial. This is the Scaled Heisenberg lattice, whose fibre is central but whose total group need not split as a direct product of groups.

A finitely presented group is one isomorphic to $F(X)/⟨⟨R⟩⟩$, where $X$ and $R$ are finite, $F(X)$ is the free group on $X$, and the double brackets mean normal closure. Equivalently, there is a surjection from a free group of finite rank onto $G$ whose kernel is normally generated by finitely many words. The universal property of the free group extends choices of generator images uniquely, and quotienting imposes the chosen relations.

For the semidirect product, choose finite group presentationsFor each $i,j$, choose an $a$-word $w_{ji}$ representing $\varphi (b_j)(a_i)$. We claim that a finite presentation of a semidirect product isAll relations hold in the desired semidirect product, so the presented group maps onto it. We must also show there are no extra elements in the kernel. For inverse conjugations, choose an $a$-word $v_{ji}$ representing $\varphi (b_j{)}^{-1}(a_i)$. Conjugating this word by $b_j$ using the displayed relations gives an $a$-word representing $\varphi (b_j)\varphi (b_j{)}^{-1}(a_i)=a_i$, so $R$ proves $b_j^{-1}{a}_i{b}_j=v_{ji}$. Thus we can move both positive and negative $b$-letters to the right, rewriting every word as $u(a)v(b)$. If its image in $N⋊H$ is trivial, projection to $H$ makes $v$ trivial using $S$; its remaining image in $N$ then makes $u$ trivial using $R$. The kernel is trivial, proving the claim and finite presentability.

For a poly-infinite cyclic group, write its subnormal series as $G=G_0▹G_1▹\cdots ▹G_n=1$ with infinite cyclic factors. Given a subgroup $K\le G$, intersect the series with $K$. The homomorphism $K\cap G_i\to G_i/G_{i+1}$ has kernel $K\cap G_{i+1}$, soEach factor is trivial or infinite cyclic; deleting repetitions gives a poly-infinite cyclic series for $K$. For an extension $1\to N\to E\to Q\to 1$ with $N,Q$ poly-infinite cyclic groups, lift the series of $Q$ to its full preimages in $E$ and append the series of $N$. The lifted quotient factors are those of $Q$, and the lower factors are those of $N$. Normality at every consecutive step is preserved. This proves closure under both subgroups and extensions.

Finite presentability follows by induction on the length of a poly-infinite cyclic series. The trivial group is finitely presented. For a nontrivial group, its first proper term $G_1$ has a shorter series and $G/G_1\cong \mathbb{Z}$. Lift $1\in \mathbb{Z}$ to an element $t\in G$. Its powers give an injective section of the quotient map, since $t^k$ maps to $k$. Hence $G\cong G_1⋊⟨t⟩$. By induction $G_1$ is finitely presented, and $⟨t⟩\cong \mathbb{Z}$ has a finite presentation. The finite presentation of a semidirect product proved above completes the induction. Thus every poly-infinite cyclic group is finitely presented.

For the final example, take the dyadic affine Baumslag-Solitar groupwhere the generator of $\mathbb{Z}$ acts on the additive group $\mathbb{Z}[1/2]$ by multiplication by two. Here is a verification of this identification, rather than an assumption about the presentation. Put $a_i=t^{-i}a{t}^i$ for $i\ge 0$. The relation gives $a_{i+1}^2=a_i$, soTheir union $A$ is abelian and is normal: conjugation by $t^{-1}$ shifts $a_i$ to $a_{i+1}$, while conjugation by $t$ shifts $a_i$ to $a_{i-1}$ for $i\ge 1$ and sends $a_0$ to $a_0^2$. Every word can consequently be written $a_i^m{t}^k$ for some $i\ge 0$ and $m,k\in \mathbb{Z}$.

Let $a$ act on the rational line by $x\mapsto x+1$ and $t$ by $x\mapsto 2x$. These affine transformations satisfy the relation, and $a_i^m{t}^k$ acts by $x\mapsto 2^{k}x+m/2^i$. It is the identity only when $k=m=0$. Thus the action is faithful and identifies $A$ with all dyadic translations. We have the required seriesBoth factors are torsion-free abelian groups, and $G$ is visibly finitely presented. But $\mathbb{Z}[1/2]$ is not finitely generated, since finite sets have bounded powers of two in their denominators. Every poly-infinite cyclic group has the maximal condition on subgroups: its factors $\mathbb{Z}$ have that condition, and Question 1 proves closure under extensions. Therefore $G$ cannot be poly-infinite cyclic. This gives a finitely presented counterexample to replacing infinite cyclic factors by arbitrary torsion-free abelian factors.

Every polycyclic group has a finite-index subgroup with a series whose nontrivial factors are infinite cyclic. This is the polycyclic finite-index lemma; it also supplies torsion-free finite-index subgroups. Such a subgroup is a poly-infinite cyclic group. Combining its series with the extension rule for Poincare duality groups makes it a duality group of dimension its Hirsch length.