= Solution
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 is
$$
\boxed{H^k(\Gamma,\mathbb Z\Gamma)=0\ (k\ne n),\qquad
H^n(\Gamma,\mathbb Z\Gamma)=D\cong\mathbb Z\ \text{as an abelian group}.}
$$
It 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 $PD^0$ <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> $P_*\to\mathbb Z$. Dualize to $P^* =\operatorname{Hom}_R(P_*,R)$. Its cohomology is zero except in degree $n$, where it is $D$. Reversing degrees therefore makes $Q_j=P_{n-j}^*$ a projective right-$R$ resolution of $D$. Finite generation and projectivity give, for every left $R$-module $V$, an isomorphism of cochain complexes
$$
\operatorname{Hom}_R(P_*,V)\cong P^*\otimes_R V.
$$
Taking cohomology and reversing the grading yields
$$
H^k(\Gamma,V)\cong\operatorname{Tor}^R_{n-k}(D,V).
$$
Since $D$ has underlying abelian <group> $\mathbb Z$, its action is a sign character $\omega:\Gamma\to\{\pm1\}$. 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 $\gamma(d\otimes v)=\omega(\gamma)d\otimes\gamma v$. Thus
$$
\boxed{H^k(\Gamma,V)\cong H_{n-k}(\Gamma,D\otimes V).}
$$
For 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 $\Gamma$-modules
$$
\mathbb Z\Gamma\cong\operatorname{Coind}_H^\Gamma(\mathbb ZH).
$$
One explicit map sends $r\in\mathbb Z\Gamma$ to the $H$-linear function $g\mapsto\operatorname{pr}_H(gr)$, where $\operatorname{pr}_H$ retains only terms in $H$. Finite index ensures that all such functions arise in this way. <Shapiro's lemma> gives the isomorphism
$$
H^k(\Gamma,\mathbb Z\Gamma)\cong H^k(H,\mathbb ZH)
$$
as 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 $\mathbb ZH$. This proves that $H$ is $PD^n$.
For the converse one needs the usual ambient torsion-free hypothesis. The standard finite-index resolution lemmas say that type $FP_\infty$ 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 $PD^n$ 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 $H=\mathbb Z\times1$. Then $H$ is index two and is $PD^1$, 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 $j\ge1$. 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 $PD^n$.
Finally, a closed aspherical <surface> has a two-dimensional contractible <universal cover> and exactly this compact-support duality, so its <surface group> is $PD^2$. 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 $PD^2$ classification theorem confirms that every integral $PD^2$ <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 $PD^2$ <groups>.
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