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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.

The Schur multiplier of a group $G$ isthe second group homology group with trivial integral coefficients. If $G\cong F/R$ is a free presentation, Hopf's formula states that

Write $I_F=\ker (\mathbb{Z}F\to \mathbb{Z})$ for the augmentation ideal. The presentation relation sequence iswhere $r[R,R]\mapsto 1\otimes (r-1)$. If $F$ is free on a set $S$, then $I_F$ is free as a left $\mathbb{Z}F$-module on the elements $s-1$, so the two modules immediately preceding $\mathbb{Z}$ are free $\mathbb{Z}G$-modules. Resolving the relation module $R/[R,R]$ by free modules and splicing produces a free resolution of $\mathbb{Z}$.

Apply the right-exact functor $\mathbb{Z}{\otimes }_{\mathbb{Z}G}-$ to this partial resolution. Its degree-two homology is the kernel ofThe coinvariant module on the left is $R/[F,R]$. On the right, the map $f-1\mapsto f[F,F]$ identifies the coinvariants with the abelianization $F/[F,F]$. The displayed map is induced by the inclusion $R\hookrightarrow F$, so its kernel isThis proves Hopf's formula.

For an abelian group $A$, the Schur multiplier of an abelian group is $M(A)\cong {\bigwedge }^{2}A$. One way to see the direct-sum rule is the degree-two Künneth theorem:A cyclic group has zero second integral group homology, whileConsequently

The Schur multiplier of a group $G$ is its second integral group homology group:

For a finite group $G$, Tate cohomology joins group cohomology and group homology through the norm map $N_G={\sum }_{g\in G}g$. For $r\ge 1$ it equals ordinary group cohomology; for $r\le -2$ it equals $H_{-r-1}(G,M)$. The middle terms are ${\hat{H}}^0(G,M)=M^G/N_{G}M$ and ${\hat{H}}^{-1}(G,M)=\ker (N_G)/I_{G}M$, where $I_{G}M$ is generated by $gm-m$. For the trivial module $\mathbb{Z}$, ${\hat{H}}^{-2}(G,\mathbb{Z})=G^{\mathrm{ab}}$. With multiplicative coefficients the norm is the product of the translates.