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?
Best Answer
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 $G$.