I am reading a paper (*[Finitely presented residually free groups](https://arxiv.org/abs/0809.3704)* by Bridson, Howie, Miller III, and Short, Theorem 5.2) where they write the following:


Since $S\_{0}$ is a group of type $FP\_{n}(\mathbb{Q})$ and there is a series $$S\_{0}\triangleleft S\_{1}\triangleleft \cdots \triangleleft S\_{l}=T$$ where each $S\_{i+1}/S\_{i}$ is finite or infinite cyclic, then by the obvious induction, $T$ is of type $FP\_{n}(\mathbb{Q})$.


I understand that if $S\_{i+1}/S\_{i}$ is finite and $S\_{i}$ of type $FP\_{n}(\mathbb{Q})$, then so is $S\_{i+1}$, because that property is inherited in finite extensions. Nevertheless, I really can't prove the same when $S\_{i+1}/S\_{i}$ is infinite cyclic.


Why is this true? Maybe it does not always hold, but in my case yes: I am working in direct products of limit groups, so $S\_{0}$ is a full subdirect product of limit groups, $S\_{0}\leq \Gamma\_{1}\times \cdots \times \Gamma\_{n}$.



Extension of $FP_{n}$ group
If $0\to A \to B \to C\to 0$ is a short exact sequence of groups and $A$, $C$ are of type $FP\_{n}$, then so is $B$. Reference: Proposition 2.2 in *Homological finiteness properties of fibre products* by Kochloukova and Ferreira Lima ([arXiv link](https://arxiv.org/abs/1611.03759)).


