False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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FALSE: Kurosh says every subgroup of a free product is free

Statement

Every subgroup of a free product is free.

Facts & Assumptions

Given: The Kurosh subgroup theorem.

[L1]

A subgroup of a free product is itself a free product of a free group together with intersections with conjugates of the factors. (Kurosh subgroup theorem)

[L2]

The factors embed in their free product. (The factor maps into a free product with amalgamation are injective)

[L3]

Every free group is torsion-free. (Free groups are torsion-free)

Refutation

technique · direct

1.1

Let $G = C_{2} * C_{3}$ and let $H = C_{2}$ be the first embedded factor, whose embedding is supplied by [L2]. In the Kurosh decomposition of $H$ , the identity double coset contributes the intersection $H \cap C_{2} = H$ .

L1L2given

2.1

The subgroup $H = C_{2}$ contains a nonidentity element of order $2$ , whereas [L3] says every free group is torsion-free. Thus $H$ is a subgroup of a free product that is not free, and the universal statement is false.

L1L3step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 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

Roger C. Lyndon and Paul E. Schupp, Combinatorial Group Theory (standard reference, not scraped)