Zero sum maps on finite abelian groups are given by permutations

Let $𝐴 = {𝑎 𝑖} 𝑛 𝑖 = 1$ be a finite abelian group of order $𝑛$ and $𝜑 : ℕ 𝑛 \to 𝐴 : 𝑖 \mapsto 𝑏 𝑖$ be a function. Then $𝜑$ satisfies $\sum 𝑛 𝑖 = 1 𝜑 (𝑖) = 0$ iff there exists a permutation

\sigma = \begin{align*} \begin{pmatrix} a_{1},\dots,a_{n} \\ c_{1},\dots,c_{n} \end{pmatrix} \end{align*} \in A!

of $𝐴$ such that $𝑐 𝑖 - 𝑎 𝑖 = 𝑏 𝑖$ .¹ comb

Proof

We first show that given a permutation
$𝜎 = (𝑎 1, \dots, 𝑎 𝑛 𝑐 1, \dots, 𝑐 𝑛)$
whose differences are $𝑏 1, \dots, 𝑏 𝑛$ , we can find another $𝜎'$ whose differences are $𝑏' 1, 𝑏' 2, 𝑏 3, \dots, 𝑏 𝑛$ such that $𝑏' 2 = 𝑏 1 + 𝑏 2 - 𝑏' 1$ , i.e. the same except in two places.

See op. cit.

develop | en | SemBr

1952. A Combinatorial Problem on Abelian Groups ↩

Zero sum maps on finite abelian groups are given by permutations

Footnotes

Graph View