CorollaryStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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The two factor images intersect exactly in the amalgamated subgroup

Statement

Inside $G\ast_KH$ , the images of $G$ and $H$ intersect exactly in their common image of $K$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

Every element of $G\ast_KH$ has a unique normal form $s_1\cdots s_nk$ relative to fixed transversals. A normal word of positive length is nonidentity. The represented group and these conclusions are independent of the chosen transversals. (Normal form theorem for free products with amalgamation).

[L2]

The canonical maps $G\to G\ast_KH$ and $H\to G\ast_KH$ are injective. (The factor maps into a free product with amalgamation are injective).

Proof

technique · direct

1.1

The image of $K$ lies in both factor images by the commuting pushout square.

givenL1L2

2.1

If the images of $g\in G$ and $h\in H$ are equal, then the normal form of $gh^{-1}$ is trivial. Normal-form uniqueness forces both factor representatives to reduce to the same length-zero element of $K$ .

step 1.1

3.1

Thus the intersection is precisely the amalgamated subgroup. The argument includes trivial $K$ and a whole-factor inclusion.

step 2.1∎

Depends on

Used by

Amalgamating C₂ inside C₄ and C₆ gives the presentation with a²=b³ Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 21 results over 8 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

George D. Torres, Combinatorial Group Theory, §2 (standard reference, not scraped)
B. H. Neumann, Lectures on Topics in the Theory of Infinite Groups, Ch. 9 (standard reference, not scraped)