Intuitive reasoning why orbits of difference sets give rise to block designs

Question

Consider the orbits of $A= \{0,3,4\}$ and $B=\{0,5,7\}$ under the action of addition mod 13. I know that the orbits of these blocks will generate the blocks of a $(13,3,1)$-design because $A$ and $B$ make up a difference-system.

I can't quite understand why this is the case? Any intuitive thoughts would be helpful.

Answer 1

Let $A=\{a_1,a_2,a_3\}$ and $B=\{b_1,b_2,b_3\}$ be three-element sets of integers modulo 13 with the property that $$ \{a_i-a_j\}_{i,j}\cup\{b_i-b_j\}_{i,j}\tag{1} $$ contains all integers modulo 13 (my interpretation of what is meant by a "difference-system"). Note that $A$ cannot be a translate of $B$, since then the two sets in $(1)$ would be the same and the union would contain at most 6 non-zero elements. Thus, there are exactly 26 sets obtainable by translating either $A$ or $B$.

I claim that this family of 26 sets forms a $(13,3,1)$-block design. The 13 total elements is clear, as is the 3 elements per block. All that is left is to show that for every pair of distinct elements $x,y$ modulo 13, there exists a unique translate of $A$ or $B$ containing $\{x,y\}$.

First let us show that there is at least one translate with this property. Since $x-y$ is nonzero mod 13, by the hypothesis in $(1)$ there exists $i,j$ such that either $x-y=a_i-a_j$ or $x-y=b_i-b_j$. (Let's assume the former case, the latter case is similar.) Then $x,y$ both belong to the translate $A+(x-a_i)$.

Finally, observe that the translate is unique, since if $x-y$ could be written as $a_i-a_j$ or $b_i-b_j$ in more than one way, then there would be fewer than 12 non-zero elements appearing in the union from $(1)$, which is a contradiction.

As for the specific sets $A$ and $B$ you have written down, they indeed form a difference-system since the non-zero differences appearing are

A:
3  4 1
10 9 12

B:
5 7 2
8 6 11