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:

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

Related facts

 * Snevily's conjecture: A related question for subsets of bigger size is an open conjecture.
 * Snevily's conjecture for cyclic groups

Facts used

 * 1) uses::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)).