In Brown's book "Cohomology of Groups" VIII Theorem 3.1, he proves a result, crediting to Serre:
Theorem. If $\Gamma$ is a torsion-free group and $\Gamma'$ is a subgroup of finite index, then $cd(\Gamma) = cd(\Gamma')$.
I want to use this to say the following
Statement: Suppose $\Gamma$ is a group with finite index torsion-free subgroups $\Gamma_1$ and $\Gamma_2$. Then $cd(\Gamma_1)= cd(\Gamma_2)$.
I have seen here and some other places that the statement seems to be true, I just cannot figure out how to prove it.
(My attempt: we can consider the subgroup $\Gamma' \leq \Gamma $ generated by $\Gamma_1,\Gamma_2$, and it does contain the $\Gamma_i$'s as finite index torsion-free subgroups, but I'm not sure why $\Gamma'$ must be torsion-free.)
The end-goal is to be able to define the virtual cohomological dimension of a group $\Gamma$, as long as $\Gamma$ contains a finite index torsion-free subgroup.
Use the intersection instead, $[\Gamma_1: \Gamma_1 \cap \Gamma_2] \le [\Gamma: \Gamma_2]$. So $\Gamma_1 \cap \Gamma_2$ is finite index in both $\Gamma_1$ and $\Gamma_2$ and, being a subgroup of a torsion-free group, is torsion-free.