Three questions on MSTD, MDTS and balanced subsets of dicyclic groups, answered in the negative

Abstract

For a subset A of a finite group one writes A+A={a₁a₂: a₁,a₂ ∈ A} and A-A={a₁a₂⁻¹: a₁,a₂ ∈ A}, and calls A sum-dominant (MSTD), balanced, or difference-dominant (MDTS) according as |A+A| exceeds, equals, or falls short of |A-A|. Writing Sd(4n, m), Dd(4n, m), Bd(4n, m) for the numbers of m-element subsets of the dicyclic group Dic(4n) of each type, Mandal and Neetu recently determined all three counts for m=2, and for m=3 with n odd, and closed their paper with three questions: whether 4 divides Sd(4n, m) for every admissible m and every n ≥ 3; whether 4n divides Sd(4n, m) for every m ≥ 2; and whether all three counts are always even. We answer all three in the negative. One cell suffices: the four-element subsets of Dic(12) split as (Sd(12, 4),Dd(12, 4),Bd(12, 4))=(126,228,141), where 126 is divisible by neither 4 nor 12 and 141 is odd. A second, independent counterexample lies in Dic(28), outside the range tabulated in the source: (Sd(28, 4),Dd(28, 4),Bd(28, 4))=(5586,7560,7329). We also compute gcd{Sd(12, m): m ≥ 2}=6; we give the complete profiles (S,D,B) of Dic(4n) for n ≤ 7 over the ranges searched, including the balanced counts, which the source does not tabulate; and we show that the failure is not uniform in n, since at Dic(20) and Dic(24) the size-four counts do satisfy both divisibility questions. Two of the three refuting values are entries of the source's own tables, and had already been published a year earlier by Neetu, Shetty and Shankar, so what is new here is the observation rather than the arithmetic; the cells at Dic(24) and Dic(28) appear in neither paper. Every theorem below is machine-checked in Lean 4.

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