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Every finitely generated torsion-free group which is a nilpotent group embeds as a discrete cocompact subgroup of a unique simply connected nilpotent real Lie group. Its dimension equals the group's integral cohomological dimension of a group. In dimension three the Lie algebra is either abelian or has a basis with a single nonzero bracket $[Y,T]=X$, giving the real Heisenberg group.

Start with the prime decomposition of a closed orientable three-manifold. Its fundamental group is a free product of the prime-summand groups. A nontrivial abelian group cannot be a free product of two nontrivial groups. The same is true of a nontrivial finitely generated nilpotent group: it has nontrivial center, whereas such a free product has trivial center. Thus all nontrivial fundamental group comes from a single prime summand; simply connected summands do not affect the argument.

A prime but nonirreducible orientable summand is $S^2\times S^1$, giving $\Gamma \cong \mathbb{Z}$. Otherwise the infinite-group prime summand is an irreducible three-manifold. The sphere theorem for three-manifolds gives ${\pi }_2=0$. Its simply connected universal cover is noncompact, so its top ordinary homology vanishes; its first and second homology also vanish by simple connectivity and the Hurewicz theorem. It is therefore acyclic and contractible by the Whitehead theorem. The summand is aspherical, and its closed orientable three-dimensional classifying space gives integral cohomological dimension of a group equal to three: the cellular resolution gives the upper bound and top Poincare duality gives the lower bound.

If $\Gamma$ is torsion-free abelian, finite generation makes it ${\mathbb{Z}}^r$. Its integral cohomological dimension of a group is $r$, as one sees from the standard $r$-torus. The aspherical case therefore has $r=3$. We obtainBoth occur, on $S^2\times S^1$ and the three-torus.

For the infinite torsion-free nilpotent case, the same reduction gives either $\mathbb{Z}$ or an aspherical group of cohomological dimension three. By the Malcev completion of a torsion-free nilpotent group, the latter is a lattice in a simply connected three-dimensional nilpotent Lie group. A nonabelian three-dimensional nilpotent Lie algebra has abelianization of dimension at least two: a nilpotent algebra generated modulo its derived algebra by one element would be generated by that element and thus be abelian. Its derived algebra is therefore one-dimensional, and nilpotence makes that line central. Hence it has a basis with a single nonzero bracket $[Y,T]=X$; the other possibility is an abelian algebra. In the nonabelian case the lattice's center is an isolated infinite cyclic subgroup, generated by $x$, and the quotient by it is ${\mathbb{Z}}^2$. Choose lifts $y,t$ of a basis of that quotient. Their commutator is $x^b$ for a nonzero integer $b$, and normal forms giveThe subgroup $A=⟨x,y⟩\cong {\mathbb{Z}}^2$ is normal and ${\Gamma }_b/A\cong \mathbb{Z}$. In its ordered basis $x,y$, conjugation by $t$ acts byThis is the unipotent torus-bundle group. Indeed, its normal forms multiply according toProjection to the $n,k$ coordinates, followed by inspection of the central $m$ coordinate, proves torsion-freeness. All commutators lie in $⟨x^b⟩$, which is central, proving nilpotence. The mapping torus of the torus automorphism with this matrix realizes every integer $b$ as a closed orientable three-manifold; $b=0$ gives ${\mathbb{Z}}^3$, while $b\ne 0$ gives the Nil cases. Thus the complete list of groups under the nilpotent hypotheses is $\mathbb{Z}$ and the groups ${\Gamma }_b$.

Here is the center argument, with its literal qualification. If a nilpotent group with abelian quotient $\Gamma /A$ were nonabelian, the last nonzero term ${\gamma }_c\Gamma$ of its lower central series would satisfyConsequently $Z(\Gamma )\cap A=1$ forces the group to be abelian. For the specific nontrivial rank-two $A$ in the printed family the hypothesis cannot actually occur: when $b\ne 0$, $x\in Z({\Gamma }_b)\cap A$; when $b=0$, the intersection is all of $A$. The implication is valid, but should not be used to invent an example satisfying this impossible extra condition.

A torsion-free finite extension of a lattice with quotient $C_4$ is realized by the quarter-turn flat three-manifold. On ${\mathbb{R}}^3$ letLet $c$ be unit translation in the third coordinate. Then $sa{s}^{-1}=b$, $sb{s}^{-1}=a^{-1}$, $sc{s}^{-1}=c$, and $s^4=c$. The lattice $\Lambda =⟨a,b,c⟩\cong {\mathbb{Z}}^3$ is normal and the quotient is $C_4$. A representative $a^m{b}^n{c}^k{s}^j$, $0\le j<4$, with $j\ne 0$ has third-coordinate displacement $k+j/4\ne 0$, so has no fixed point; for $j=0$ a nonidentity element is a nonzero translation. This proves torsion-freeness. The action is properly discontinuous, cocompact and orientation preserving, so it gives a closed orientable flat three-manifold with precisely the required extension.

For faithful linear holonomy on the maximal translation lattice, the holonomy groups of closed orientable flat three-manifolds areThe cyclic examples replace the quarter-turn by a planar lattice rotation of order $m=2,3,4,6$ and a translation of $1/m$ along its axis. The square and triangular plane lattices realize these orders; the trace condition $2\cos (2\pi /m)\in \mathbb{Z}$ explains the crystallographic restriction. The remaining case is obtained fromTheir squares translate by one in the first and second directions, and $(\alpha \beta {)}^2$ translates by minus one in the third. Their linear parts generate $C_2\times C_2$. The three nontranslation cosets have half-integral displacement along their respective fixed axes, so every element acts without a fixed point. These constructions realize every member of the orientable three-dimensional Euclidean holonomy classification.

The maximality/faithfulness restriction is essential. Taken literally as a question about arbitrary finite group quotients, the wording has more answers: $\mathbb{Z}$ has every finite cyclic quotient, and ${\mathbb{Z}}^3$ has, for example, quotient $C_5$. An exact criterion for the literal extension question is as follows. Choose an action $Q\to \mathrm{Aut}(A)$ and an extension class $e\in H^2(Q,A)$, where $A=\mathbb{Z}$ or ${\mathbb{Z}}^3$. For every subgroup $⟨q⟩$ of prime order $p$, requireTo prove this, lift $q$ to $g$. Its $p$th power is in $A^q$, and replacing $g$ by $ag$ changes that power by the displayed norm of $a$. Thus a lift of order $p$ exists exactly when this restriction is zero. Any torsion element has a prime-order power, so the condition is both necessary and sufficient. For an oriented flat realization in rank three the linear action must also preserve orientation. This distinguishes a complete extension criterion from the finite faithful-holonomy list.

First correct the group-ring condition in the PDF. For an infinite group, a left-invariant element of $\mathbb{Z}\Gamma$ would have constant coefficients on the whole group; finite support then forces it to be zero. Hence $H^0(\Gamma ,\mathbb{Z}\Gamma )=0$. For a Poincare duality group of positive dimension, the correct condition isIt must be accompanied by the finiteness condition: the trivial module has a finite-length resolution by finitely generated projective modules. The module $D$ has a right group action, possibly an orientation-sign action. The degree-zero exception only applies to the trivial $P{D}^0$ group, not simultaneously to positive-dimensional groups.

Here is how the duality follows. Put $R=\mathbb{Z}\Gamma$ and take a length-$n$ projective resolution $P_{\ast }\to \mathbb{Z}$. Dualize to $P^{\ast }={\mathrm{Hom}}_R(P_{\ast },R)$. Its cohomology is zero except in degree $n$, where it is $D$. Reversing degrees therefore makes $Q_j=P_{n-j}^{\ast }$ a projective right-$R$ resolution of $D$. Finite generation and projectivity give, for every left $R$-module $V$, an isomorphism of cochain complexesTaking cohomology and reversing the grading yieldsSince $D$ has underlying abelian group $\mathbb{Z}$, its action is a sign character $\omega :\Gamma \to \{\pm 1\}$. Twisting the resolution by this character identifies the right side with group homology having coefficient module $D{\otimes }_{\mathbb{Z}}V$, with diagonal left action $\gamma (d\otimes v)=\omega (\gamma )d\otimes \gamma v$. ThusFor an orientable Poincare duality group, $D$ is trivial and this is untwisted Poincare duality. The natural isomorphism can equivalently be realized by cap product with the group fundamental class. The reversed finite projective complex explains why the top group-ring condition produces duality for every coefficient module, rather than just matching integral Betti numbers.

Now let $H$ have finite index in $\Gamma$. There is an isomorphism of left $\Gamma$-modulesOne explicit map sends $r\in \mathbb{Z}\Gamma$ to the $H$-linear function $g\mapsto {\mathrm{pr}}_H(gr)$, where ${\mathrm{pr}}_H$ retains only terms in $H$. Finite index ensures that all such functions arise in this way. Shapiro's lemma gives the isomorphismas abelian groups. If $\Gamma$ is a Poincare duality group, restricting its finite projective resolution to $H$ preserves projectivity and finite generation: the group ring is finite free over $\mathbb{Z}H$. This proves that $H$ is $P{D}^n$.

For the converse one needs the usual ambient torsion-free hypothesis. The standard finite-index resolution lemmas say that type $F{P}_{\infty }$ is preserved under finite-index passage, and that a torsion-free group of finite virtual cohomological dimension has the same integral cohomological dimension as a finite-index subgroup. Thus if $H$ is $P{D}^n$ and $\Gamma$ is torsion-free, $\Gamma$ has a finite resolution by finitely generated projective modules of length $n$. The preceding explicit coinduction and Shapiro calculation transfers exactly the required group-ring cohomology, and proves the finite-index invariance of Poincare duality for torsion-free groups.

Without that hypothesis the claimed converse is false. Take $\Gamma =\mathbb{Z}\times C_2$ and $H=\mathbb{Z}\times 1$. Then $H$ is index two and is $P{D}^1$, while $\Gamma$ contains torsion. Its integral cohomological dimension of a group is infinite: restriction of any putative finite projective resolution to $C_2$ would contradict the nonzero cohomology $H^{2j}(C_2,\mathbb{Z})=\mathbb{Z}/2$ for every $j\ge 1$. Also, orientability need not ascend: the torus subgroup of the Klein-bottle group has trivial orientation action, whereas the full group has a nontrivial one. The invariant theorem concerns the possibly twisted notion of $P{D}^n$.

Finally, a closed aspherical surface has a two-dimensional contractible universal cover and exactly this compact-support duality, so its surface group is $P{D}^2$. Conversely the algebraic condition forces the same local duality dimension and, in the orientable case, a nonsingular skew pairing on first cohomology, the familiar algebra of a closed orientable surface. Infinite-index subgroups behave like noncompact surfaces rather than closed ones. These are strong reasons to expect a surface group; in fact the $P{D}^2$ classification theorem confirms that every integral $P{D}^2$ group is the fundamental group of a closed aspherical surface. This includes the torus and, with orientation twist, the Klein bottle; the sphere and projective plane do not give $P{D}^2$ groups.