TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Normal form theorem for free products with amalgamation

Statement

Every element of $G *_{K} H$ has a unique normal form $s_{1} \dots s_{n} k$ relative to fixed transversals. A normal word of positive length is nonidentity. The represented group and these conclusions are independent of the chosen transversals.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

Let $K$ be embedded in $G$ and $H$ as in def-free-product-with-amalgamation. By def-axiom-of-choice, choose left-coset transversals $S_{G}, S_{H}$ containing the identity. A normal word is $s_{1} \dots s_{n} k,$ where $n \in N$ (def-natural-numbers), $k \in K$ , every $s_{j}$ is a nonidentity representative from $S_{G}$ or $S_{H}$ , and consecutive representatives come from different factors. Length zero means the word is just $k$ . The written form depends on the transversals. (Transversal normal-form data for an amalgamated free product).

[L2]

With fixed transversal data, every element of $G$ and $H$ acts by a permutation on normal words. The two actions agree on $K$ , inverses act inversely, and with the library's composition convention one has $P_{x y} = P_{x} \circ P_{y}$ . (Factor elements act consistently by permutations on amalgamated normal words).

[L3]

Let $G = ⟨ X ∣ R ⟩$ and $H = ⟨ Y ∣ S ⟩$ with disjoint generators, and let $f, h$ embed $K$ . If $T$ generates $K$ and words $u_{t} (X), v_{t} (Y)$ represent $f (t), h (t)$ , then $G *_{K} H ≅ ⟨ X ⊔ Y ∣ R \cup S \cup {u_{t} v_{t}^{- 1} : t \in T} ⟩ .$ (A free product with amalgamation has the factor presentations plus the amalgamating relations).

[L4]

Given homomorphisms $f : K \to G$ and $h : K \to H$ as in def-group-homomorphism, a pushout is a group $P$ with homomorphisms $i_{G} : G \to P$ and $i_{H} : H \to P$ such that $i_{G} \circ f = i_{H} \circ h$ , and such that every compatible pair $u : G \to Q$ , $v : H \to Q$ factors through a unique $w : P \to Q$ with $w \circ i_{G} = u$ and $w \circ i_{H} = v$ . The maps $f, h$ need not be injective. (Pushouts of group homomorphisms).

Proof

technique · direct

1.1

The compatible factor permutations give, by the pushout presentation, an action of $G *_{K} H$ on the set of normal words. For a factor product $w$ , applying $P_{w}$ to the length-zero word performs the deterministic normal-form rewrite of $w^{- 1}$ ; since inversion is a bijection of the group, every element has such a form.

givenL1L2L3L4

2.1

The permutation attached to a normal word sends the length-zero word to the deterministic normal-form rewrite of its inverse. The resulting inversion-normalisation map is an involution: invert the represented factor product again and repeat the uniquely determined transversal rewrites. It preserves syllable length, since multiplying a nontrivial transversal representative by an element of $K$ cannot put it in $K$ . Hence a positive-length normal word cannot represent the identity, and equality of two represented elements forces equality of their inverse normal forms and then of the original words.

step 1.1

3.1

Thus existence and uniqueness hold, including the length-zero elements of $K$ .

step 2.1

4.1

Changing transversals gives another group with the same pushout universal property; the unique factor-compatible isomorphism identifies the two descriptions, so the group and its conclusions do not depend on the choices.

step 3.1∎

Depends on

Used by

Dependency tree · two levels

14 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

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)