We say that a group *G* is in the class *Fq* if there is a CW-complex which is a *BG* (that is, which has fundamental group *G* and contractible universal cover) and which has finite *q*-skeleton. Thus *F0* contains all groups, *F1* contains exactly the finitely generated groups, *F2* the finitely presented groups, and so forth.


My question: For a fixed *q* ≥ 3, is it possible to decide, from a finite presentation of a group *G*, whether *G* is in *Fq* or not? I would assume not, but am not having much luck proving it.


One approach would be to prove that, if *G* is a group in *Fq* and *H* is a finitely presented subgroup, then *H* ∈ *Fq* as well. This would make being in *Fq* a Markov property, or at least close enough to make it undecidable.


Henry Wilton's comment below makes it clear that being *Fq* is not even quasi-Markov, so the above idea won't work. I still suspect that "*G* ∈ *Fq*" is not decidable, but now my intuition is from Rice's theorem:



> 
> If $\mathcal{B}$ is a nonempty set of computable functions with nonempty complement, then no algorithm accepts an input *n* and decides whether *φn* is an element of $\mathcal{B}$.
> 
> 
> 


It seems likely to me that something similar is true of finite presentations and the groups they define.


John Stillwell notes below that this can't be true for a number of questions involving the abelianization of G. This wouldn't affect the Rips construction/1-2-3 theorem discussion below if the homology-sphere idea works, since those groups are all perfect.


Any thoughts?



Is any interesting question about a group G decidable from a presentation of G?
It seems to me that the analogue of Rice's theorem fails for finitely presented
groups $G$ because of questions like: is the abelianization of $G$ of rank 3?
The rank of the abelianization of any finitely presented $G$ can be computed
by reducing the abelianization to normal form, so this (slightly) interesting
question can be decided from the presentation of $G$.


