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The fundamental group of a graph with trivial groups is free
Statement
If every vertex group and edge group of a graph of groups is trivial, then its fundamental group is free. If the underlying graph is finite, the rank equals the number of geometric edges outside a maximal subtree.
Facts & Assumptions
Given: A graph of groups with all vertex and edge groups trivial.
The graph-of-groups fundamental group acts on its Bass-Serre tree, and the quotient graph is the original underlying graph. (The fundamental group acts without inversions on its Bass-Serre tree)
A group acting freely without inversions on a tree is free. (A group acting freely without inversions on a tree is free)
Proof
By [L1], the graph-of-groups fundamental group acts on its Bass-Serre tree. Because every stabilizer is trivial, this action is free and without inversions.
Applying [L2] to the action from step 1.1 shows that the fundamental group is free. If the quotient graph is finite, a maximal subtree uses all vertices and all but the non-tree geometric edges, so the free basis from the previous corollary has one generator for each such edge.
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Sources
- C. Loh, Geometric Group Theory: An Introduction (2015 course version) (standard reference, not scraped)
- John Meier, Groups, Graphs and Trees (standard reference, not scraped)