Snevily's conjecture for subsets of size two

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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.

Related facts

Facts used

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


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)).