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Nov 29, 2022 at 11:51 comment added HJRW ... (cont'd). What is true is that every pushout canonically defines a graph of groups -- just replace every vertex and edge group with its image in the pushout. But, as in Rylee's example, the vertex groups might become trivial when you do this.
Nov 29, 2022 at 11:50 comment added HJRW @AntoineLabelle: What you're talking about is arbitrary pushouts of groups. As you say, they can be useful -- eg every quotient is a pushout, and of course we like to study quotients of groups. But the most basic result of the theory of graphs of groups -- specifically, the fact that the vertex groups embed -- fails for arbitrary pushouts. So we need a name for this "stronger" theory, and the name for that is graphs of groups.
Nov 29, 2022 at 3:55 comment added Antoine Labelle Why is that particularly problematic? For example, the tensor product of nontrivial rings can sometimes be trivial, but that doesn't mean it's not useful to define and study the tensor product in full generality.
Nov 29, 2022 at 2:17 history answered Robbie Lyman CC BY-SA 4.0