In the definition of a graph of groups it is assumed that the maps from the edge groups to the vertex groups of injections, however in what follows I will also be interested in the case where the maps from the edge groups to the vertex groups are **n**ot necessarily **i**njective. I will call these n.i. graphs of groups. There is already a question and some helpful answers about such n.i. graphs of groups [here](https://mathoverflow.net/questions/293183/graphs-of-groups-with-homomorphisms-not-necessarily-injective).


As noted in that linked question, given a connected n.i. graph of groups $\mathcal{G}$ we can go ahead and define the fundamental group $\pi\_1(\mathcal{G})$ just as in the case of a normal graph of groups (either combinatorially as Serre does in "Trees" or topologically by constructing the appropriate graph of spaces and taking the fundamental group of its total space as in "Topological methods in group theory" Scott and Wall).


Is there a fintely presented group $G$ such that $G = \pi\_1(\mathcal{G})$ for some n.i. graph of groups $\mathcal{G}$ with all of the vertex and edge groups of $\mathcal{G}$ finite, but $G$ can not be decomposed as the fundamental group of a (good ol' classic edge-map injective) graph of groups where all of the vertex and edge groups are finite?


What about the same question but where we replace "finite" with "finitely generated free" or "finitely generated abelian" or "finitely generated free abelian"??



Realizing groups as the fundamental group of graphs of groups allowing non-injective maps?
Let $\mathcal C$ be a class of groups closed under homomorphic images. Then any ni-graph of groups whose vertex and edge groups belong to $\mathcal C$ has the same fundamental group as an ordinary graph of groups with vertex and edge groups in $\mathcal C$ as per @HJRW's answer to [this question](https://mathoverflow.net/questions/293183/graphs-of-groups-with-homomorphisms-not-necessarily-injective).
Namely if $\mathcal G$ is your ni-graph of groups, keep the same underlying graph but replace $G\_x$ by its image $\overline{G}\_x$ in $\pi\_1(\mathcal G)$ for each $x$ a vertex or edge to get a usual graph of groups with the same fundamental group. This covers the case of finite groups and finitely generated abelian groups.


For finitely generated free groups or finitely generated free abelian groups, you cannot replace you ni-graph of groups by an ordinary graph of groups of the same sort. Let $G=\langle X\mid R\rangle$ be a finitely presented group with $X,R$ finite. Then build an ni-graph of groups with a single edge between two vertices and one vertex group trivial and the other free on $X$. Let the edge group be free on $R$ and map it into the trivial group the unique way and into the free group on $X$ by extending the inclusion of $R$. Then $G$ is the fundamental group of this graph of groups and so any finitely presented group can be obtained in this way.


Similarly, if $A$ is any finitely generated abelian group we can give it a finite presentation $A=\langle X\mid R\rangle$ as an *abelian group* and now we imitate the above to write $A$ as the fundamental group of the ni-graph of groups with trivial vertex group and a free abelian group on $X$ as the other vertex group and the free abelian group on $R$ as the edge group and map into the two vertex groups as in the previous paragraph.


