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A vanishing group-ring coefficient sum pairs off opposite-signed equal labels

Statement

Let π be a group and let g1,…,gr∈π and ε1,…,εr∈{±1} satisfy ∑j=1rεj[gj]=0 in Z[π], or ∑j=1rεj[gj]=[h] for a single element h∈π. If r≥2, then there are indices j1≠j2 with gj1=gj2 and εj1=−εj2. In particular, in the single-monomial case with r≥2 the multiset of signed labels contains an opposite-signed pair of equal labels.

Facts & Assumptions

Given: A group π, a finite list g1,…,gr∈π of elements and signs ε1,…,εr∈{±1}, together with the equality ∑j=1rεj[gj]=0, or the equality ∑j=1rεj[gj]=[h] for a single element h∈π.

[F1]

The classes [g], g∈π, form a Z-basis of Z[π]: the group ring is the free left Z-module on the set π, and every element of Z[π] has a unique expression ∑g∈Frg[g] with F⊆π finite and rg∈Z, so two such expressions are equal if and only if they have the same coefficient at every label; in particular the expression of 0 has every coefficient 0, and the expression of a single basis vector [h] has coefficient 1 at h and coefficient 0 at every other label (The group ring R[G] of finitely supported formal R-linear combinations of group elements, The group ring R[G] is a unital R-algebra with basis G, and each g∈G is a unit of R[G]).

Proof

1.1F1algebra

Group the terms of the sum by label: for each g∈π set kg:=∑j: gj=gεj∈Z, a finite sum that is nonzero only for the finitely many occurring labels, so that ∑j=1rεj[gj]=∑g∈πkg[g] is the expansion of the left-hand side in the basis of [F1]. By the uniqueness of that expansion, the equality ∑j=1rεj[gj]=0 holds exactly when kg=0 for every g∈π, and the equality ∑j=1rεj[gj]=[h] holds exactly when kh=1 and kg=0 for every g≠h.

2.1step 1.1contradiction

Assume no two indices carry equal labels with opposite signs, so that for every occurring label g all terms with gj=g share one sign and kg is ± the number of occurrences of g, hence a nonzero integer. If ∑j=1rεj[gj]=0, then step 1.1 forces kg to vanish for every label, contradicting the nonzero coefficient of each occurring label; therefore in the zero-sum case some pair of indices has equal labels and opposite signs.

2.2step 1.1algebra

Assume no two indices carry equal labels with opposite signs and consider the single-monomial case ∑j=1rεj[gj]=[h] with r≥2. By step 1.1 every label g≠h must have kg=0, and under the assumption every occurring label has a nonzero coefficient, so no label other than h occurs and all r terms carry the label h. Then kh=∑j=1rεj=r−2m, where m counts the indices with εj=−1, and the equation kh=1 with r≥2 gives an odd r≥3 and 1≤m≤r−1; hence some index has sign +1 and some index has sign −1, both with label h, so an opposite-signed pair of equal labels exists in the single-monomial case as well.

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 settle the zero-sum and the single-monomial case respectively, so under the stated hypotheses and r≥2 there are always indices j1≠j2 with gj1=gj2 and εj1=−εj2. The argument used only the basis expansion of Z[π], no property of π beyond it and no choice principle.

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