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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

1 Answers

One Answer

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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