LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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.

Factor elements act by mutually inverse permutations on reduced syllable words

Statement

For each $i\in I$ and $g\in G_i$ , left multiplication at the first syllable defines a permutation $P_{i,g}$ of the set of reduced words. One has $P_{i,g}^{-1}=P_{i,g^{-1}}$ and $P_{i,gh}=P_{i,g}\circ P_{i,h}$ , so $g\mapsto P_{i,g}$ is a group homomorphism.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

For groups as in def-group, a syllable is a tagged pair $(i,g)$ with $i\in I$ and $g\in G_i\setminus\{e_i\}$ . A reduced syllable word is a finite list of syllables, indexed by a natural length as in def-natural-numbers, in which adjacent tags differ. The empty list is allowed. At a concatenation seam, adjacent syllables from the same factor are multiplied and an identity result is deleted; this elementary reduction is repeated until the seam is reduced. (Reduced syllable words in a family of groups).

[L2]

For every set $X$ , the triple $(\operatorname{Sym}(X), \circ, \mathrm{id}_X)$ of def-symmetric-group is a group (def-group); the inverse of a permutation $f$ is its inverse function $f^{-1}$ . If $X$ contains three distinct elements $a$ , $b$ , $c$ , then $\operatorname{Sym}(X)$ is not abelian: the transpositions $\tau = (a\,b)$ and $\rho = (b\,c)$ satisfy $\tau \circ \rho \ne \rho \circ \tau$ . ( $\operatorname{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

[L3]

Let $(M,\cdot,e)$ and $(M',\cdot',e')$ be monoids (def-semigroup-and-monoid). A monoid homomorphism from $M$ to $M'$ is a function $f : M \to M'$ such that - (H1) $f(x \cdot y) = f(x) \cdot' f(y)$ for all $x, y \in M$ ; - (H2) $f(e) = e'$ . Let $G$ and $G'$ be groups (def-group). A group homomorphism from $G$ to $G'$ is a function $f : G \to G'$ satisfying (H1) alone: $f(xy) \;=\; f(x)\, f(y) \qquad \text{for all } x, y \in G .$ Condition (H2) is not imposed for groups because it follows: a group homomorphism automatically satisfies $f(e) = e'$ and $f(x^{-1}) = f(x)^{-1}$ (lem-group-homomorphism-basic-properties). For monoids it does not follow and must be assumed, which is why the two definitions differ. A homomorphism from a structure to itself is an endomorphism. The identity map of $M$ is a monoid homomorphism, and a composite of monoid homomorphisms is one, since $(g \circ f)(xy) = g(f(x)f(y)) = g(f(x))\,g(f(y))$ and $(g \circ f)(e) = g(e') = e''$ ; the same computation, without the second clause, shows a composite of group homomorphisms is a group homomorphism. (Monoid homomorphism and group homomorphism).

Proof

technique · direct

1.1

Define $P_{i,e_i}$ to be the identity map, since $(i,e_i)$ is not a syllable and prepending it would leave a word that is not reduced. For $g\ne e_i$ , define $P_{i,g}$ by prepending $(i,g)$ when the word is empty or begins in another factor; when it begins $(i,h)$ , replace that syllable by $(i,gh)$ and delete it if $gh=e_i$ . Every value is again a reduced word.

givenL1L2L3

2.1

Let $g\ne e_i$ , so also $g^{-1}\ne e_i$ , and let $w$ be reduced. Three seam cases exhaust the definition. (a) $w$ empty or with first tag other than $i$ : $P_{i,g}(w)=(i,g)w$ , which begins $(i,g)$ , so $P_{i,g^{-1}}$ replaces that syllable by $(i,g^{-1}g)=(i,e_i)$ and deletes it, returning $w$ . (b) $w=(i,h)w'$ with $gh\ne e_i$ : $P_{i,g}(w)=(i,gh)w'$ , and $P_{i,g^{-1}}$ replaces $(i,gh)$ by $(i,g^{-1}gh)=(i,h)$ , kept because $h\ne e_i$ , returning $w$ . (c) $w=(i,h)w'$ with $gh=e_i$ , that is $h=g^{-1}$ : $P_{i,g}(w)=w'$ , and $w'$ is empty or has first tag other than $i$ because $w$ is reduced, so $P_{i,g^{-1}}(w')=(i,g^{-1})w'=(i,h)w'=w$ . Exchanging $g$ and $g^{-1}$ gives the other composite, so $P_{i,g^{-1}}$ is a two-sided inverse of $P_{i,g}$ ; with $P_{i,e_i}=\mathrm{id}$ this makes every $P_{i,g}$ a permutation of the reduced words and $P_{i,g}^{-1}=P_{i,g^{-1}}$ [L2].

step 1.1L1L2

3.1

For $P_{i,gh}=P_{i,g}\circ P_{i,h}$ both sides are immediate when $g=e_i$ or $h=e_i$ , so let $g,h\ne e_i$ and take $w$ reduced. (a) $w$ empty or with first tag other than $i$ : $P_{i,h}(w)=(i,h)w$ , and $P_{i,g}$ sends it to $(i,gh)w$ when $gh\ne e_i$ and to $w$ when $gh=e_i$ , which is $P_{i,gh}(w)$ in both subcases. (b) $w=(i,a)w'$ with $ha\ne e_i$ : $P_{i,h}(w)=(i,ha)w'$ , and $P_{i,g}$ sends it to $(i,gha)w'$ or, when $gha=e_i$ , to $w'$ ; $P_{i,gh}(w)$ splits on the same product $(gh)a=gha$ and gives the same word. (c) $w=(i,a)w'$ with $ha=e_i$ : $P_{i,h}(w)=w'$ , empty or with first tag other than $i$ , so $P_{i,g}(w')=(i,g)w'$ ; and $gha=g\ne e_i$ , so $P_{i,gh}$ replaces $(i,a)$ by $(i,g)$ and also gives $(i,g)w'$ . Hence $g\mapsto P_{i,g}$ satisfies (H1) of [L3] into the symmetric group of [L2], and is a group homomorphism.

step 2.1L1L2L3∎

Depends on

Used by

Reduced syllable words form the free product of a family of groups Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 21 results over 11 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

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)