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

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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.

Free groups on disjoint bases freely multiply to the free group on their union

Statement

For pairwise disjoint sets $X_{i}$ , the free product of the free groups $F (X_{i})$ is a free group on $⨆_{i} X_{i}$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

A free group on a set $X$ is a group $F (X)$ together with a map $i : X \to F (X)$ such that, for every group $G$ and every function $u : X \to G$ , there is a unique group homomorphism $\hat{u} : F (X) \to G$ satisfying $\hat{u} \circ i = u .$ The reduced-word construction supplies such a group; the construction and its universal property are established in thm-reduced-words-form-the-free-group. When no ambiguity arises, $x \in X$ is identified with its image $i (x)$ . (Free group on a set of generators).

[L2]

If $(F, i)$ and $(F^{'}, i^{'})$ are free groups on the same set $X$ , then there is a unique group isomorphism $ϕ : F \to F^{'}$ such that $ϕ \circ i = i^{'} .$ (Free groups on the same set are uniquely isomorphic compatibly with their generators).

[L3]

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).

[L4]

Any two free products of the same family are connected by a unique isomorphism commuting with every canonical factor map. (Free products are unique up to a unique factor-compatible isomorphism).

Proof

technique · direct

1.1

A function from the disjoint union $⨆_{i} X_{i}$ to a group $H$ is exactly a family of functions $X_{i} \to H$ .

givenL1L2L3L4

2.1

Freeness extends each member uniquely to a homomorphism $F (X_{i}) \to H$ , and free-product universality extends that family uniquely to one homomorphism from $*_{i} F (X_{i})$ .

step 1.1

3.1

Thus the free product has the universal property of $F (⨆_{i} X_{i})$ , and uniqueness gives the isomorphism. Empty bases and an empty family are included.

step 2.1∎

Depends on

Used by

Dependency tree · two levels

8 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)