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

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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.
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The word-quotient group $W (X) / \sim$ satisfies the universal property of the free group on $X$

Statement

For every set $X$ , the group $F_{w o r d} (X) = W (X) / \sim$ together with $i_{w o r d} (x) = [x]$ is a free group on $X$ in the sense of Free group on a set of generators.

Facts & Assumptions

Given: A set $X$ , a group $G$ , and a function $u : X \to G$ .

[L1]

$F_{w o r d} (X)$ is a group under $[w] [v] = [w v]$ , with identity the empty-word class $[ε]$ and $[a_{1} \dots a_{n}]^{- 1} = [a_{n}^{- 1} \dots a_{1}^{- 1}]$ ( $F_{w o r d} (X)$ is a group under $[w] [v] = [w v]$ ).

[F1]

A group homomorphism $f : G \to H$ satisfies $f (x y) = f (x) f (y)$ for all $x, y \in G$ , and consequently preserves the identity and inverses (Monoid homomorphism and group homomorphism).

[F2]

In a group, for every $x$ there is $y$ with $y x = e = x y$ (Group and abelian group).

[F3]

A free group on $X$ is a group with a map from $X$ for which every function from $X$ to a group extends uniquely to a group homomorphism (Free group on a set of generators).

Proof

technique · constructive

1.1

Extend $u$ to formal letters by $\tilde{u} (x) = u (x)$ and $\tilde{u} (x^{- 1}) = u (x)^{- 1}$ , and for $w = a_{1} \dots a_{n}$ define $E (w) = \tilde{u} (a_{1}) \dots \tilde{u} (a_{n})$ , with $E (ε) = e_{G}$ .

F2givenconstruct

2.1

An elementary insertion or cancellation changes this product only by inserting or deleting an adjacent factor $u (x) u (x)^{- 1}$ or $u (x)^{- 1} u (x)$ , which equals $e_{G}$ ; hence one elementary move leaves $E (w)$ unchanged.

F2step 1.1

3.1

A finite sequence of elementary moves therefore preserves evaluation, so $\hat{u} ([w]) : = E (w)$ is well-defined on equivalence classes.

step 2.1construct

4.1

For words $w, v$ , one has $E (w v) = E (w) E (v)$ , so [L1] and [F1] show that $\hat{u}$ is a homomorphism; moreover $\hat{u} ([x]) = u (x)$ , so it extends $u$ .

L1F1step 3.1

5.1

If $h : F_{w o r d} (X) \to G$ is any homomorphism with $h ([x]) = u (x)$ , then [F1] gives $h ([x^{- 1}]) = h ([x]^{- 1}) = u (x)^{- 1}$ . For $w = a_{1} \dots a_{n}$ , the class $[w]$ is the ordered product of its one-letter classes, so [F1] forces $h ([w]) = \tilde{u} (a_{1}) \dots \tilde{u} (a_{n}) = \hat{u} ([w])$ ; hence $h = \hat{u}$ .

L1F1step 4.1

6.1

The homomorphism of step 4.1 exists for every $G$ and $u$ , and step 5.1 makes it unique; by [F3], $(F_{w o r d} (X), i_{w o r d})$ is a free group on $X$ , including when $X$ is empty.

F3step 4.1step 5.1discharge-construct∎

Depends on

Used by

Dependency tree · two levels

13 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

Nicholas Touikan, An Introduction to Combinatorial and Geometric Group Theory, §1.3 (standard reference, not scraped)
Richard Elman, Lectures on Abstract Algebra, §18 (standard reference, not scraped)

The word-quotient group W(X)/∼W(X)/∼ satisfies the universal property of the free group on XX