= Solution
A <nilpotent group> has <lower central series> $\gamma_1G=G$, $\gamma_{j+1}G=[G,\gamma_jG]$ 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 $\gamma_jG/\gamma_{j+1}G$ 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 $H\triangleleft G$. 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 gives
$$
1\longrightarrow H\longrightarrow P\longrightarrow\mathbb Z\longrightarrow1,
$$
where $H$ has a poly-infinite cyclic series of length two. Thus $H\cong\mathbb Z\rtimes\mathbb Z$, 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 $\langle a,b\mid bab^{-1}=a^{-1}\rangle$. Its characteristic subgroup $\langle a,b^2\rangle\cong\mathbb Z^2$ is the centralizer of its <commutator subgroup> $\langle a^2\rangle$. 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 gives
$$
\boxed{1\longrightarrow\mathbb Z^2\longrightarrow P_0\longrightarrow\mathbb Z\longrightarrow1,\qquad [G:P_0]<\infty.}
$$
This extension splits: any lift $t$ of the cyclic generator supplies a section. Therefore $P_0=\mathbb Z^2\rtimes_B\mathbb Z$ for some $B\in GL_2(\mathbb Z)$.
In additive notation on $\mathbb Z^2$, $[t,v]=(B-I)v$. All its other internal <commutators> vanish, so induction gives
$$
\gamma_{j+1}P_0=(B-I)^j\mathbb Z^2\quad(j\geq1).
$$
Consequently
$$
\boxed{P_0\text{ is nilpotent}\iff(B-I)^2=0
\iff\operatorname{tr}B=2\text{ and }\det B=1.}
$$
This 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 $\mathbb Z^2\rtimes_{-I}\mathbb Z$ 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 give
$$
(x,y,z)(x',y',z')=(x+x',y+y',z+z'+xy'),\qquad
(x,y,z)^{-1}=(-x,-y,-z+xy).
$$
These 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 to
$$
(x+ka,\ y+kb,\ z+kc+xkb).
$$
Thus $(x\bmod k,y\bmod k)$ 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 $z\mapsto z+kc$. Hence
$$
U\times(\mathbb R/k\mathbb Z)\longrightarrow p^{-1}(U),\qquad
((x,y),[z])\longmapsto(x,y,z)N_k
$$
is 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 (mathematics)> is its <abelianization>. Let $a=(k,0,0)$, $b=(0,k,0)$ and $c=(0,0,k)$. Every element is uniquely $a^mb^nc^\ell$, since its third coordinate is $k^2mn+k\ell$. The <commutator> formula is
$$
[(x,y,z),(x',y',z')]=(0,0,xy'-x'y),
$$
so $c$ 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 $N_k'=\langle c^k\rangle$. The homomorphism $(x,y,z)\mapsto(x/k,y/k,z/k\bmod k)$ 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 $k$. Therefore
$$
\boxed{H_1(\mathrm{Nil}^3/N_k;\mathbb Z)\cong N_k/N_k'\cong\mathbb Z^2\oplus\mathbb Z/k.}
$$
For $k=1$ 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>.
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