Uniqueness (upto isomorphism) of 2-designs

Question

Suppose I have a $2-(v,k,\lambda)$ design with $v=6,\,k=3,\, \lambda=2$ , how do I show that it is unique upto isomorphism? I have already shown that two blocks must intersect at either one or two points.

Answer 1

Basically you have to hack it out. Call the points $A,\ldots,F$. Consider the blocks through $A$. Taking out $A$ you get the edges of a graph on $\{B,\ldots,F\}$. This graph is regular of degree $2$ so is a $5$-cycle, so we may assume we have blocks

$ABC\quad ACD\quad ADE\quad AEF\quad ABF$.

Now for the five remaining blocks I claim we cannot have a block like $BCD$ with three consecutive vertices on the cycle $BCDEF$. If we did, the only possible block through $C$ and $E$ would be $CEF$. This leaves only five blocks

$BCE\quad CDF\quad DEB\quad EFC\quad FBD$

which must the the last five blocks.