Filtered colim of F-groups

A group G is said to have a property F if there exists a finite aspherical CW-complex of which it is the fundamental group (according to wikipedia).

question: is there a full characterization of groups that are obtained as a filtered colimits of F-groups?

Thanks.

Edit: May be more simple (?) question: Is any torsion free group a filtered colimit of F-groups?

MathOverflow Asked by Ofra on January 21, 2021

Here is an answer to the simpler question, "Is any torsion free group a filtered colimit of F-groups?" (Or, just to be clear, let me rephrase the question as, "Is every torsion free group a filtered colimit of F-groups?")

The answer is no: Thompson's group $$F$$ is torsion free but I claim it is not a filtered colimit of F-groups. Let $$G$$ be the colimit of some filtered system of groups $$(G_i)$$, so as explained in https://math.stackexchange.com/questions/362502/finitely-generated-subgroups-of-direct-limits-of-groups every finitely presented subgroup of $$G$$ is isomorphic to a subgroup of $$G_j$$ for some $$j$$ (note that this is assuming what you call "filtered colimit" is the same as what is called "direct limit" in this link; from doing a little research it seems this is the case). Now suppose $$G=F$$, which since $$F$$ itself is finitely presented tells us that $$F$$ is isomorphic to a subgroup of $$G_j$$ for some $$j$$. Lastly, suppose all the $$G_i$$ are F-groups, so we get that $$F$$ is isomorphic to a subgroup of some F-group. This is a contradiction since $$F$$ contains $$mathbb{Z}^infty$$ but no F-group can contain $$mathbb{Z}^infty$$. We conclude $$F$$ is not a filtered colimit of any system of F-groups.

Note that this argument is not really that specific to $$F$$; it shows that any finitely presented subgroup of a filtered colimit of F-groups must be isomorphic to a subgroup of an F-group. In particular the class "finitely presented groups embeddable into a filtered colimit of F-groups" coincides with the class "finitely presented groups embeddable into an F-group".

Correct answer by Matt Zaremsky on January 21, 2021

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