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FALSE: the quotient graph determines the acting group without stabilizer data
Statement
If two groups act on trees with the same quotient graph, then the acting groups must be isomorphic.
Facts & Assumptions
Given: The Bass-Serre structure theorem.
A tree action is recovered from the full quotient graph of groups, including the vertex and edge stabilizers and the boundary monomorphisms. (Bass-Serre structure theorem)
A one-loop graph of groups has the corresponding HNN extension as its fundamental group. (A one-loop graph of groups gives an HNN extension)
A graph-of-groups fundamental group acts on its Bass-Serre tree with the original underlying graph as quotient. (The fundamental group acts without inversions on its Bass-Serre tree)
Every free group is torsion-free. (Free groups are torsion-free)
Every vertex group embeds in its graph-of-groups fundamental group. (Vertex groups embed in the fundamental group of a graph of groups)
Refutation
Take a one-loop quotient graph. Giving it trivial vertex and edge groups produces the fundamental group by [L2]. Giving the same loop vertex group and trivial edge group produces . By [L3], each fundamental group acts on its Bass-Serre tree with that same one-loop quotient graph.
The first group is free and hence torsion-free by [L4], while [L5] embeds the order- vertex subgroup in the second, so they are not isomorphic. Thus the quotient graph alone does not determine the acting group.
Depends on
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
19 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Jean-Pierre Serre, Trees (standard reference, not scraped)