Difference set

Remarks

When $G$ is abelian and written additively, the condition is about differences $d_1 - d_2$. Counting the $k(k - 1)$ pairs $(d_1, d_2)$ with $d_1 \ne d_2$ gives $k(k-1) = \lambda (v - 1)$. Example: $D = \set{1, 2, 4}$ in $\Z/7\Z$. The six differences are $2 - 1 = 1$, $1 - 2 = 6$, $4 - 1 = 3$, $1 - 4 = 4$, $4 - 2 = 2$, $2 - 4 = 5$, so each nonzero residue occurs once, and $D$ is a $(7, 3, 1)$ difference set. Non-example: $D = \set{0, 1, 2}$ in $\Z/7\Z$ is not a difference set, since $1 = 1 - 0 = 2 - 1$ occurs twice while $3$ does not occur. If $D$ is a $(v, k, \lambda)$ difference set with $k < v$, the translates $Dg = \set{dg : d \in D}$ for $g \in G$ are the blocks of a symmetric $2$-$(v, k, \lambda)$ block design.