Rank of a free Abelian subgroup gives a lower bound for vcd

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?

• Hint: cohomological dimension is monotonic with respect to subgroups. This should be in Brown. – Moishe Kohan Apr 19 at 0:17
• 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 $H'$? – yshen Apr 20 at 10:29
• Of course a finite index subgroup of $G$ need not be a finite index subgroup of $H$, but you can always intersect it with $H$. – Moishe Kohan Apr 20 at 12:57
• 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...? – yshen Apr 20 at 14:48
• The intersection has finite index in $H$ and hence, has the same vcd as $H$. – Moishe Kohan Apr 20 at 14:52