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

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The factor maps into a free product with amalgamation are injective

Statement

The canonical maps $G\to G\ast_KH$ and $H\to G\ast_KH$ are injective.

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]

A group homomorphism is injective if and only if its kernel is trivial. For a group homomorphism $f:G\to H$ , $f$ is injective exactly when $\ker f=\{e_G\}$ . (A group homomorphism is injective if and only if its kernel is trivial).

Proof

technique · direct

1.1

A nonidentity factor element rewrites either as a nontrivial length-zero element of $K$ or as a normal word with one nonidentity transversal syllable.

givenL1L2

2.1

Neither form is the identity by the normal-form theorem, so each canonical map has trivial kernel and is injective. This includes trivial factors and the case where $K$ is a whole factor.

step 1.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 37 results over 13 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)