# Snevily's conjecture for subsets of size two

## Statement

Suppose $G$ is an odd-order abelian group and $A = \{ a_1, a_2 \}$ and $B = \{ b_1, b_2 \}$ are (not necessarily disjoint) subsets of $G$ of size two. Then, one of these is true:

$a_1 + b_1 \ne a_2 + b_2, \qquad a_1 + b_2 \ne a_2 + b_1$.

## Facts used

1. kth power map is bijective iff k is relatively prime to the order

## Proof

Given: A finite Abelian group $G$ of odd order, subsets $A = \{ a_1, a_2 \}$ and $B = \{ b_1, b_2 \}$ of $G$.

To prove: Either $a_1 + b_1 \ne a_2 + b_2$ or $a_1 + b_2 \ne a_2 + b_1$.

Proof: Suppose equality holds in both cases. Then, subtracting the two equations, we get:

$(a_1 + b_1) - (a_1 + b_2) = (a_2 + b_2) - (a_2 + b_1)$.

Rearranging this, we get:

$2(b_1 - b_2) = 0$.

Since $b_1 \ne b_2$, $b_1 - b_2 \ne 0$, and since the group is Abelian of odd order, its double is also therefore nonzero (using fact (1)).