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A universal finite abelian carrier bound

Abstract

The complete original support family forces a quadratic lower bound in every finite abelian carrier.

Theorem 1.1 (Every finite abelian realization needs at least 2m(m-1) elements).

Lean statement: D5/S3/Combinatorics/Sequenceability/OriginalFiniteAbelianCarrierLowerBound.finite_abelian_carrier_card_lower_bound

Proof. Machine-checked in Lean as D5/S3/Combinatorics/Sequenceability/OriginalFiniteAbelianCarrierLowerBound.finite_abelian_carrier_card_lower_bound (✓ std3). ∎

Source. Repository-derived.

Commentary.

Let G be any finite abelian group, m any natural number at least three, and f any labeling of the whole original Coordinate m. Assume the q label has additive order exactly m. Assume the complete original zero-support classification: for every finite original coordinate subset C, nonemptiness together with a zero label sum is equivalent to C being precisely {q,a_i,b_i} for an original pair index i. Then G has at least 2m(m-1) elements. This theorem assumes no label injectivity, cyclicity, oddness, quotient exponent restriction, or selected-support replacement for the full classification.

Put H equal to the subgroup generated by f(q), K = G/H, and A_i equal to the quotient class of f(a_i). The existing original-support separation lemmas give A_i nonzero, distinct values at distinct indices, and A_i different from -A_j at distinct indices. For m at least four, the existing all-signs triple lemma gives A_i-A_j-A_k nonzero whenever i, j and k are pairwise distinct. The relation f(a_j)+f(b_j)=-f(q) identifies the class of each b_j with -A_j. Thus the signed-triple exclusion used here is derived from the complete original classification.

Write r=m-1 for the number of pair indices. If none of the selected A_i is self-negative, all 2r elements A_i and -A_i are distinct. Hence the quotient already has at least 2r elements. This branch uses no restriction on other involutions of K and no assumption that K is cyclic.

Otherwise select h=A_o=-A_o, which is nonzero and has additive order exactly two. Consider the further quotient K/. The classes of the A_i remain injective. A collision at distinct indices would give A_i-A_j=zero or h, because the subgroup generated by h consists of exactly zero and h. The first alternative contradicts the original injectivity. In the second, i=o forces A_j=zero and j=o forces A_i=2*h=zero. With neither index equal to o, the equation A_i-A_j-A_o=zero is the forbidden signed triple. This handles every collision, including those involving the selected self-negative point.

There are therefore at least r elements in K/. Since has exactly two elements, K has at least 2r elements. In both branches, H has exactly m elements and card(G)=card(K)card(H), proving the stated lower bound. At m=three the already established generic bound m(m+1) equals 2m*(m-1), so that boundary case needs no triple argument and no new numerical theorem.

This lower bound and the existing affine binary construction determine an infinite family of exact minima among all finite abelian carriers with the original hypotheses. If d is at least one and m-1=2^d, then m is odd and the affine construction has exactly m2^(Nat.clog 2 (m-1)+1)=m2^(d+1)=2m(m-1) elements. It has the same exact q order and full original zero-support classification, so it attains the universal lower bound. No exact minimum is asserted for parameters whose predecessor is not a power of two, and no literature-priority claim is made.

References