Let $G$ be any group. The cohomological dimension (cd) of $G$ is the smallest integer $n$ such that $\mathbb{Z}$ admits a projective resolution of length $n$ over the group ring $\mathbb{Z}G$. Serre proved that the cohomological dimensions of all the torsion-free subgroups of finite index are the same, and the common cohomological dimension is called the virtual cohomological dimension (vcd) of $G$.

An equivalent definition of cd($G$) is the smallest integer $n$ such that $H^i(G, -) = 0$ for $i >n$. (see Brown's Cohomology of Groups, p.185)

Let $H \leq G$ be a free Abelian group of rank $n$. Then the classifying space $BH$ is the $n$-fold torus $T^n = S^1 \times \cdots \times S^1$, which is a $K(H,1)$ with $H^n(T^n,\mathbb{Z}) \cong \mathbb{Z} \neq 0$ and all higher cohomology groups vanish. Therefore cd$(H) = n$. (The argument can be found in Brown p.185.) Being free Abelian implies that $H$ is torsion free but not necessarily of finite index in $G$. I want to show that $\text{vcd}(G) \geq \text{cd}(H) = n$. How should I proceed?

  • $\begingroup$ Hint: cohomological dimension is monotonic with respect to subgroups. This should be in Brown. $\endgroup$ – Moishe Kohan Apr 19 at 0:17
  • $\begingroup$ Thanks for the hint... assume that there exists a torsion free subgroup of finite index $H'$ in $G$, do I know that $H$ is a subgroup of $H'$? $\endgroup$ – yshen Apr 20 at 10:29
  • $\begingroup$ Of course a finite index subgroup of $G$ need not be a finite index subgroup of $H$, but you can always intersect it with $H$. $\endgroup$ – Moishe Kohan Apr 20 at 12:57
  • $\begingroup$ Then the intersection is a subgroup of $H$ and has cohomological dimension smaller than $n$, which doesn't help with establishing $n$ as the lower bound for vcd...? $\endgroup$ – yshen Apr 20 at 14:48
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    $\begingroup$ The intersection has finite index in $H$ and hence, has the same vcd as $H$. $\endgroup$ – Moishe Kohan Apr 20 at 14:52

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