We say that a group *G* is in the class *F _{q}* 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

*F*contains all groups,

_{0}*F*contains exactly the finitely generated groups,

_{1}*F*the finitely presented groups, and so forth.

_{2}My question: For a fixed *q* ≥ 3, is it possible to decide, from a finite presentation of a group *G*, whether *G* is in *F _{q}* 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 *F _{q}* and

*H*is a finitely presented subgroup, then

*H*∈

*F*as well. This would make being in

_{q}*F*a Markov property, or at least close enough to make it undecidable.

_{q}Henry Wilton's comment below makes it clear that being *F _{q}* is not even quasi-Markov, so the above idea won't work. I still suspect that "

*G*∈

*F*" is not decidable, but now my intuition is from Rice's theorem:

_{q}If $\mathcal{B}$ is a nonempty set of computable functions with nonempty complement, then no algorithm accepts an input

nand decides whetherφis an element of $\mathcal{B}$._{n}

~~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?