ExampleConstruction: AI-generatedVerification: 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.

The canonical surjection from a free product to the direct product of its factors

Example

For groups $G, H$ , the factor maps $g \mapsto (g, e)$ and $h \mapsto (e, h)$ induce a canonical surjection $π : G * H ↠ G \times H .$ Every cross-commutator lies in $\ker π$ , and if $g, h$ are nonidentity then that commutator is nonidentity in the free product.

Facts & Assumptions

Given: The objects and hypotheses in the example.

[L1]

For a family $(G_{i})_{i \in I}$ , a free product is a group $F$ with homomorphisms $ι_{i} : G_{i} \to F$ in the sense of def-group-homomorphism, such that for every group $H$ and every family of homomorphisms $f_{i} : G_{i} \to H$ , there is a unique homomorphism $f : F \to H$ satisfying $f \circ ι_{i} = f_{i}$ for all $i$ . It is denoted $*_{i \in I} G_{i}$ . Injectivity of the maps $ι_{i}$ is not part of this definition. (The free product of an arbitrary family of groups).

[L2]

Every element of $*_{i \in I} G_{i}$ has a unique reduced syllable expression. The identity is represented by the empty word, and no nonempty reduced word represents the identity. (Normal form theorem for free products).

[L3]

Let $G$ and $H$ be groups. Their external direct product has underlying set $G \times H : = {(g, h) : g \in G, h \in H}$ and componentwise operation $(g, h) (g^{'}, h^{'}) : = (g g^{'}, h h^{'}) .$ The fact that this operation makes $G \times H$ a group, with the indicated identity and inverses, is proved in thm-external-direct-product-is-a-group. Until that result is used, this definition introduces only the set and its componentwise binary operation. (The external direct product $G \times H$ with componentwise multiplication).

[L4]

For groups $G$ and $H$ , the componentwise operation of def-external-direct-product-of-groups makes $G \times H$ a group. Its identity is $(e_{G}, e_{H})$ , and $(g, h)^{- 1} = (g^{- 1}, h^{- 1}) .$ Moreover the coordinate maps $π_{G} (g, h) = g$ and $π_{H} (g, h) = h$ are group homomorphisms. ( $G \times H$ is a group with identity $(e_{G}, e_{H})$ , coordinatewise inverses, and homomorphic coordinate projections).

[L5]

Let $G$ be a group. For $g, h \in G$ , their commutator is $[g, h] : = g h g^{- 1} h^{- 1} .$ This convention is fixed throughout; some sources use its inverse. By the inverse laws of lem-group-inverse-laws, one has $[g, h]^{- 1} = h g h^{- 1} g^{- 1} = [h, g]$ . The commutator subgroup, or derived subgroup, is the subgroup generated by all commutators: $[G, G] : = ⟨ {[g, h] : g, h \in G} ⟩ .$ The generated subgroup notation is that of def-generated-subgroup. (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

Verification

technique · direct

1.1

Free-product universality gives $π$ , and every $(g, h) = (g, e) (e, h)$ lies in its image, so it is surjective.

givenL1L2L3L4L5

2.1

The two factor images commute in $G \times H$ , hence $[g, h]$ maps to the identity.

step 1.1

3.1

For nonidentity $g \in G$ , $h \in H$ , the word $g h g^{- 1} h^{- 1}$ is a nonempty reduced word, so normal form makes it nonidentity. Thus the kernel is generally nontrivial.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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