Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11
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Factor elements act by mutually inverse permutations on reduced syllable words

Statement

For each i∈I and g∈Gi, left multiplication at the first syllable defines a permutation Pi,g of the set of reduced words. One has Pi,g−1=Pi,g−1 and Pi,gh=Pi,g∘Pi,h, so g↦Pi,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∈I and g∈Gi∖{ei}. 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 (Sym⁡(X),∘,idX) 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 Sym⁡(X) is not abelian: the transpositions τ=(a b) and ρ=(b c) satisfy τ∘ρ≠ρ∘τ. (Sym⁡(X) is a group under composition, and it is non-abelian whenever X has at least three distinct elements).

[L3]

Let (M,⋅,e) and (M′,⋅′,e′) be monoids (def-semigroup-and-monoid). A monoid homomorphism from M to M′ is a function f:M→M′ such that - (H1) f(x⋅y)=f(x)⋅′f(y) for all x,y∈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→G′ satisfying (H1) alone: f(xy)  =  f(x) f(y)for all x,y∈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∘f)(xy)=g(f(x)f(y))=g(f(x)) g(f(y)) and (g∘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 Pi,ei to be the identity map, since (i,ei) is not a syllable and prepending it would leave a word that is not reduced. For g≠ei, define Pi,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=ei. Every value is again a reduced word.

givenL1L2L3
2.1

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

step 1.1L1L2
3.1

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

step 2.1L1L2L3∎

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