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My question is about the computation the fundamental group of graph of groups:

First let me give a reference for my question:

In the page 14 of Groups acting on graphs by Dunwoody and Dicks.

It says that

The fundamental group of graphs of groups can be obtained by successively performing one free product with amalgamation for each edge in the maximal subtree and then one HNN extension for each edge not in the maximal subtree.

I cannot see this procedure in the following example.

Suppose that our graph $Y$ is the complete graph with three vertices $K_3$ and the vertex groups are $A,B,C$ and the edge groups are $E,F,D$.

Jivid
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  • What are you looking for exactly? What is giving you trouble from seeing the procedure in your particular case? The description of what you should do is pretty clear. Are you familiar with Seifert–van Kampen theorem? –  Oct 19 '17 at 13:44
  • @PaulPlummer let me explain. When you fix a a path of length 2, then you you have a free product with amalgamation for one edge. Then it is not clear for me the rest. so probably you remove that edge and replace the old group of the vertex with a free product with amalgamation obtained and then we have a new free product with amalgamation for the remained single edge. Am I correct? – Jivid Oct 19 '17 at 14:11

1 Answers1

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The trick is that at each step of the procedure, the group computed in the previous step is treated as a single vertex group in the next step.

So, for example, in the left-most diagram $1-2-3$, one first computes the fundamental group of the one-edge subgraph $1-2$, which is $A*_EB$. That becomes a single vertex group in the middle diagram, the other vertex group being $C$ which is the group label on the original vertex 3, and the edge group being $F$ which is the group labelling the original edge $2-3$. The edge-to-vertex homomorphism $F \to A*_EB$ in the middle diagram is defined to be the composition $$F \mapsto B \mapsto A*_EB $$

One way to think of what has happened from the left diagram to the middle diagram is that the $1-2$ edge in the left diagram has been collapsed to form a vertex in the middle diagram, not that the edge has been removed in any way as your comment suggests.

Lee Mosher
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  • I found this answer trying to understand essentially the same thing. I have some question and would be great if you can clarify it to me. (1) in this example you gave, the fundamental group will be $(A \ast_E B) \ast_F C$? (2) The book mention vertex not in maximal subtree. How it works? For example. In diagram $1-2-3$ suppose that we can a closed triangle. We can chose a maximal subgroup formed exactly by $1-2-3$. The edge $1-3$ is not in the maximal subtree. How we do it in this case? $(A \ast_E B) \ast_F C$ will be a vertex with a loop edge labelled by the original label of $1-3$? – Lucas Mar 09 '23 at 21:52
  • I apologize if I was not clear. – Lucas Mar 09 '23 at 21:53
  • In general, asking new questions in the comment of an old answer is a poor use of this site; only the author of this answer (me) will ever see it. Much better would be to post a new question. – Lee Mosher Mar 10 '23 at 01:59
  • Ok. I create a new question. – Lucas Mar 10 '23 at 03:01