ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck 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 free group on one generator is isomorphic to $(Z, +)$

Example

Let $X = {x}$ be a one-element set. Every free group $(F, i)$ on $X$ is isomorphic to $(Z, +)$ , by the isomorphism carrying $i (x)$ to $1$ .

Facts & Assumptions

Given: The word-quotient free group $F_{w o r d} ({x})$ .

[L1]

Every class in $W (X) / \sim$ contains exactly one reduced word (Every class in $W (X) / \sim$ contains exactly one reduced word).

[L2]

If $g$ has infinite order, then for $j, k \in Z$ , $g^{j} = g^{k}$ implies $j = k$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

[L3]

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

[L4]

The word-quotient group $W (X) / \sim$ , with $x \mapsto [x]$ , is a free group on $X$ (The word-quotient group $W (X) / \sim$ satisfies the universal property of the free group on $X$ ).

[L5]

The cyclic subgroup generated by $g$ is exactly the set of integer powers of $g$ : $⟨ g ⟩ = {g^{n} : n \in Z}$ ( $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian).

[L6]

Integer powers satisfy $g^{m + n} = g^{m} g^{n}$ for all $m, n \in Z$ (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

Verification

technique · constructive

1.1

A reduced word on ${x, x^{- 1}}$ cannot contain both letters, since a change from one to the other creates an adjacent inverse pair; hence every reduced word is uniquely $x^{n}$ for $n \geq 0$ or $x^{- n}$ for $n \geq 1$ , with the empty word corresponding to exponent $0$ .

L1given

2.1

Thus $[x]$ generates the group, and no positive power $[x]^{n}$ is the identity because its reduced representative $x^{n}$ is nonempty; therefore $[x]$ has infinite order.

L1step 1.1

3.1

Since $[x]$ generates by step 2.1, [L5] makes every element of $F_{w o r d} ({x})$ equal to $[x]^{k}$ for some $k \in Z$ , and [L2] applied to the infinite-order element $[x]$ makes that exponent unique; so $θ ([x]^{k}) : = k$ is a well-defined injection, it is surjective because $k = θ ([x]^{k})$ , and [L6] makes it a homomorphism. Construct $θ : F_{w o r d} ({x}) \to (Z, +)$ as this isomorphism, which carries $[x]$ to $1$ .

L2L5L6step 2.1construct

4.1

By [L4] the word-quotient model is a free group on $X$ , so for any free group $(F, i)$ on $X$ the isomorphism $ϕ : F \to F_{w o r d} ({x})$ of [L3] satisfies $ϕ (i (x)) = [x]$ ; then $θ \circ ϕ$ is an isomorphism $F \to (Z, +)$ carrying $i (x)$ to $1$ .

L3L4step 3.1discharge-construct∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

44 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

John McKernan, Presentations and Groups of Small Order, Lecture 12 (standard reference, not scraped)

The free group on one generator is isomorphic to (Z,+)(Z,+)