TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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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 $F_{\mathrm{word}}(X)=W(X)/{\sim}$ together with $i_{\mathrm{word}}(x)=[x]$ 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]

$F_{\mathrm{word}}(X)$ is a group under $[w][v]=[wv]$ , with identity the empty-word class $[\varepsilon]$ and $[a_1\cdots a_n]^{-1}=[a_n^{-1}\cdots a_1^{-1}]$ ( $F_{\mathrm{word}}(X)$ is a group under $[w][v]=[wv]$ ).

[F1]

A group homomorphism $f:G\to H$ satisfies $f(xy)=f(x)f(y)$ 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 $yx=e=xy$ (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 $\widetilde u(x)=u(x)$ and $\widetilde u(x^{-1})=u(x)^{-1}$ , and for $w=a_1\cdots a_n$ define $E(w)=\widetilde u(a_1)\cdots\widetilde u(a_n)$ , with $E(\varepsilon)=e_G$ .

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 $\widehat u([w]):=E(w)$ is well-defined on equivalence classes.

step 2.1construct

4.1

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

L1F1step 3.1

5.1

If $h:F_{\mathrm{word}}(X)\to G$ is any homomorphism with $h([x])=u(x)$ , then [F1] gives $h([x^{-1}])=h([x]^{-1})=u(x)^{-1}$ . For $w=a_1\cdots a_n$ , the class $[w]$ is the ordered product of its one-letter classes, so [F1] forces $h([w])=\widetilde u(a_1)\cdots\widetilde u(a_n)=\widehat u([w])$ ; hence $h=\widehat u$ .

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], $(F_{\mathrm{word}}(X),i_{\mathrm{word}})$ is a free group on $X$ , including when $X$ is empty.

F3step 4.1step 5.1discharge-construct∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 22 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)/{\sim}W(X)/∼ satisfies the universal property of the free group on XXX