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ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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A nonreduced word deleted by its repeated prefix reflection, and an exchange step

Example

Let S={s,t} with m(s,t)=3, so that W=I2(3) is the dihedral group of order 6 in which st has order 3 (The rank-two block computation, exact dihedral orders, the signed reflection action, and ambient reducedness (4), Reduced words and lengths in a finite dihedral group (1)).

  1. Exchange. The word (s,t) is a reduced expression of st with ℓ(st)=2. Since s⋅st=t has length 1<2, the exchange theorem (Length parity, exchange, two-letter deletion, and faithfulness of the signed reflection action (2)) predicts s⋅st=t=s1⋯s1^⋯s2, the deletion of the first letter; and indeed s st=t.
  2. A nonreduced word. In the word (s,t,s,t) the prefix reflections are r1=s, r2=sts, r3=t, r4=s; thus r1=r4. Deleting the first and the last letter gives the word (t,s) with value ts, and indeed stst=(st)2=(st)−1=ts, so the deleted word is an expression of the same element: stst=ts.
  3. Consequences. The word (s,t,s,t) is nonreduced: ℓ(stst)=ℓ(ts)=2<4, in agreement with the length formula ℓ((st)2)=min⁡(4,2)=2 of Reduced words and lengths in a finite dihedral group (3). The reflection set of the element is Φ(stst)={sts,t}, of cardinality 2=ℓ(stst), and s∉Φ(stst) because s=r1=r4 occurs an even number of times; deleting a different pair, e.g. the letters s at positions 1 and 3, does not preserve the value: t⋅t=1≠ts.

Facts & Assumptions

Given: The Coxeter matrix on S={s,t} with m(s,t)=3; the presented group W with its length ℓ of Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups; the reflection set T, the prefix reflections, the sign-change sets Φ(w) and the exact order of st from The rank-two block computation, exact dihedral orders, the signed reflection action, and ambient reducedness; the exchange and deletion statements of Length parity, exchange, two-letter deletion, and faithfulness of the signed reflection action; and the length table for I2(3) of Reduced words and lengths in a finite dihedral group.

[F1]

The rank-two block computation, exact dihedral orders, the signed reflection action, and ambient reducedness (1),(4),(6): "σs2=idE for every s∈S; each σs is invertible, and ℓ(s)=1"; "Consequently, for any distinct s,t∈S, one has s≠t in W and st has order exactly m(s,t) in W (infinite when m(s,t)=∞).", so here st has order 3; "(a) If ri=rj for some i<j, then s1⋯si^⋯sj^⋯sk=w: the two letters si,sj can be deleted"; "(b) (−1)n(r)=:η(r,w) depends only on w and r"; and for a reduced word "(c) ... the set Φ(w):={r1,…,rk}={r∈T:η(r,w)=−1} is independent of the reduced expression chosen, with #Φ(w)=ℓ(w)", where ri=s1⋯si−1sisi−1⋯s1 are the prefix reflections and n(r)=#{i:ri=r}.

[F2]

Length parity, exchange, two-letter deletion, and faithfulness of the signed reflection action (2): "Let w=s1⋯sk be a reduced expression and let s∈S satisfy ℓ(sw)=k−1. Then sw=s1⋯si^⋯sk for some i∈{1,…,k}"; and (3): "If the word (s1,…,sk) in S is not reduced, then there are i<j with s1⋯si^⋯sj^⋯sk=s1⋯sk."

[F3]

Reduced words and lengths in a finite dihedral group (3): for 0≤k≤m one has "ℓ((st)k)=min⁡(2k, 2(m−k))", so with m=3 and k=2 this reads ℓ((st)2)=min⁡(4,2)=2.

[F4]

Powers gn: natural exponents in a monoid and integer exponents in a group, with g0=e: gk is the k-fold product with g0=1 and g−k=(g−1)k, so g⋅g−1=1 and (st)−1=t−1s−1=ts in W, where s−1=s and t−1=t because s2,t2 are relators of the presentation of Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups.

[F5]

Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups: the relators are s2, t2 and (st)3, and ℓ(w) is the least length of a word in S representing w.

Verification

technique · direct computation in $I_2(3)$, with the prefix reflections read off from the definition and the general exchange and deletion statements applied to this word
1.1F1F2F3

The exchange step. The word (s,t) has value st and length 2, and ℓ(st)=2 by [F3] (the case m=3, k=1 gives min⁡(2,4)=2), so (s,t) is a reduced expression of st. In W one has s2=1, so s⋅st=s2t=t; and ℓ(t)=1 by [F1]. Hence ℓ(s⋅st)=1=2−1 and [F2] (2) applies to the reduced word (s1,s2)=(s,t) with the letter s: there is i∈{1,2} with s⋅st=s1⋯si^⋯s2. Since s⋅st=t and the two deletion words are s1^s2=t and s1s2^=s, only i=1 gives the value t, so s⋅st=s1^s2=t: the predicted deletion is the deletion of the first letter.

1.2F1F4F5

The prefix reflections and the repeated one. Compute the prefix reflections of (s,t,s,t) from ri=wi−1siwi−1−1 with w0=1, w1=s, w2=st, w3=sts: r1=s; r2=s t s=sts; r3=(st) s (st)−1=sts⋅ts=ststs=(st)2s=ts⋅s=t, where (st)−1=ts and (st)2=(st)−1 because st has order 3 [F1, F4]; and r4=(sts) t (sts)=s t s t s t s=(st)3s=s, using (st)3=1 and s2=t2=1 [F5]. Hence r1=r4=s while r2=sts and r3=t.

1.3F1F4

The deletion and the value identity. Since r1=r4, [F1] (6)(a) with i=1, j=4 gives stst=s1^tss4^=ts. Independently, (st)2=(st)−1 because st has order 3 [F1], and (st)−1=t−1s−1=ts [F4]; hence stst=(st)2=ts, confirming that deleting the first and the last letter of (s,t,s,t) preserves the value.

2.1F1F3F5step 1.2

The element and its reflection set. By step 1.3 the word (s,t,s,t) represents stst=ts and has length 4, while ℓ(stst)=ℓ(ts)=2 by [F3] with m=3, k=2 and k=1; hence the word is not reduced. By [F1] (6)(b) the function η(r,stst)=(−1)n(r) is expression-independent, so it may be computed from the word (s,t,s,t) with the prefix reflections of step 1.2: of these r1=r4=s occurs twice and r2=sts, r3=t occur once each, so Φ(stst)={r∈T:η(r,stst)=−1}={sts, t} has cardinality 2=ℓ(stst), in agreement with the cardinality forced for any reduced expression by [F1] (6)(c); in particular s∉Φ(stst), and the two-letter deletion licensed by [F1] (6)(a) is the one deleting the equal reflections r1=r4, not an arbitrary pair of equal letters. The last point: deleting the letters at positions 1 and 3 of (s,t,s,t) leaves the word (t,t) with value t2=1 [F5], and 1≠ts since ℓ(1)=0≠2=ℓ(ts); so that deletion does not preserve the value.

3.1step 1.1step 1.2step 1.3step 2.1∎

Collected. The example exhibits the exchange step of [F2] (2) in complete detail (step 1.1), a nonreduced word whose two equal prefix reflections r1=r4 license the deletion of its first and last letter (steps 1.2, 1.3), and the resulting expression-independent sign-change set Φ(stst)={sts,t} together with a failed deletion of an unequal-reflection pair (step 2.1).

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