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A one-segment graph of groups gives an amalgamated free product
Statement
If a graph of groups has two vertices joined by one geometric edge and that edge lies in the chosen maximal subtree, then its fundamental group is the amalgamated free product of the two vertex groups over the edge group.
Facts & Assumptions
Given: A one-segment graph of groups with vertex groups and edge group .
A free product with amalgamation is the pushout of the two injective edge maps and . (Free products with amalgamation along monomorphisms)
The relative fundamental group is obtained from the path group by killing the chosen tree edge. (The fundamental group of a graph of groups relative to a maximal tree)
Proof
Because the unique geometric edge belongs to the maximal subtree, [L2] kills the edge symbol. The only remaining generators are the two vertex groups, and the only remaining cross relation is that the two images of the edge group agree.
That is exactly the pushout presentation named in [L1], so the fundamental group of the one-segment graph of groups is .
Depends on
Used by
Dependency tree · two levels
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Sources
- Roger C. Lyndon and Paul E. Schupp, Combinatorial Group Theory (standard reference, not scraped)
- Jean-Pierre Serre, Trees (standard reference, not scraped)