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Jun 3, 2021 at 19:07 comment added Benjamin Steinberg A standard graph of torsion free groups is torsion free. Torsion is always conjugate to a vertex group
Jun 3, 2021 at 16:23 vote accept user101010
Jun 3, 2021 at 16:23 comment added user101010 Oh, on second thought, I guess you can just take a group with no nontrivial free subgroups, for example.
Jun 3, 2021 at 12:50 comment added user101010 Thanks this is very helpful! Maybe this is trivial (sorry I'm new to these things) - how do you construct such a f.p. group $G$ and prove that it can not be $\pi_1(\mathcal{G})$ for some graph of groups $\mathcal{G}$ with free group edges/vertices? I don't really know how to prove the analogous result in the f.g abelian case either.
Jun 2, 2021 at 20:03 history edited Benjamin Steinberg CC BY-SA 4.0
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Jun 2, 2021 at 20:02 comment added Benjamin Steinberg OK. I guess it is. I kind of felt that the question is open ended. I'll revise
Jun 2, 2021 at 19:58 comment added Stefan Witzel How is this not a complete answer?
Jun 2, 2021 at 16:37 history answered Benjamin Steinberg CC BY-SA 4.0