TITLE: A group of type F that is an extension of type F-by-type F
QUESTION [11 upvotes]: Let us first recall that a group of type $F$ is a group admitting a compact classifying space. 
Let $K$ and $Q$ be groups of type $F$. Consider the family $\mathcal{G}(K, Q)$ consisting of groups $G$ of type $F$ for which there exists a short exact sequence
$$
1 \to K \to G \to Q \to 1. 
$$
Is it true that all groups in $\mathcal{G}(K, Q)$ have the same cohomological dimension?
If not, please give me a counterexample.

REPLY [6 votes]: Here is a partial answer. Theorem 5.5 of Homological Dimension of Discrete Groups by Bieri implies that the cohomological dimension of $G$ is the sum of the cohomological dimensions of $K$ and $Q$ provided $H^n(K,ZK)$ is free abelian where n is the cohomological dimension of $K$. He also only needs $K$ and $Q$ of type FP.