$cd_{\mathbb Q}$ is the cohomological dimension on rational coefficients. Notice that if we replace $\mathbb Q$ with $\mathbb Z$ in $cd_{\mathbb Q}$ the statement is certainly false (think about $\mathbb Z$ and $\mathbb Z / 2 \mathbb Z$).

Suppose that $cd_{\mathbb Q}(G) < \infty$. By a theorem of Fel'dman (see Proposition 2.5 in "Normal subgroups in duality groups and in groups of cohomological dimension 2" of Bieri) if we also assume that $cd_{\mathbb Q}(G/\mathbb Z) < \infty$, we have: $$ cd_{\mathbb Q}(G) = cd_{\mathbb Q}(\mathbb Z) + cd_{\mathbb Q}(G / \mathbb Z) = 1 + cd_{\mathbb Q}(G / \mathbb Z). $$

It follows that $cd_{\mathbb Q}(G/\mathbb Z) < cd_{\mathbb Q}(G)$ if and only if $cd_{\mathbb Q}(G/\mathbb Z) < \infty$.

So the question is: under which condition we can be sure that $G/\mathbb Z$ has finite cohomological dimension? what about $G$ being a poincare group? a CAT(0) group? $\mathbb Z$ being (un)distorted? (those are just random conditions that I put as examples).

Are there known counterexamples (i.e. $cd_{\mathbb Q}(G/ \mathbb Z) = \infty$)?