= Solution
A <Steiner system> $S(l,m,n)$ is an $n$-point set together with $m$-element blocks such that every $l$-element subset is contained in exactly one block. On $X=H\cup H'$, define the following six-element blocks using the incidence extensions of $\Theta$.
Take the two blocks $H$ and $H'$. For each <duad> $d$ of $H$ and each <duad> $d'$ in the <syntheme> $\Theta(d)$, take
$$
B(d,d')=d\cup(H'\setminus d'),\qquad
X\setminus B(d,d')=(H\setminus d)\cup d'.
$$
These give forty-five blocks of type $(2,4)$ and forty-five of type $(4,2)$. Finally, for every corresponding partition pair $A\mid A^c$ and $B\mid B^c$, take all four unions
$$
A\cup B,\quad A\cup B^c,\quad A^c\cup B,\quad A^c\cup B^c.
$$
There are ten partition pairs and forty blocks of type $(3,3)$. Distinct indexing data give distinct blocks within each family, and different types have different intersection sizes with $H$. Thus the total is
$$
\boxed{2+45+45+40=132\text{ distinct blocks}.}
$$
To prove the defining property, fix a five-element subset $Y$ and set $r=|Y\cap H|$.
If $r=5$ or $0$, only $H$ or $H'$ can contain it. If $r=1$, write $Y\cap H=\{a\}$ and $Y\cap H'=C'$ with $|C'|=4$. A containing block must have type $(2,4)$ and its omitted <duad> is $d'=H'\setminus C'$. The total $T_a$ has exactly one <syntheme> containing $d'$; the other total containing that <syntheme> determines a unique second point $b$. Thus the unique block is $B(\{a,b\},d')$.
If $r=4$, write $Y\cap H=C$ and $Y\cap H'=\{a'\}$. A containing block must have type $(4,2)$, with omitted <duad> $d=H\setminus C$. Exactly one <duad> $d'\in\Theta(d)$ contains $a'$, so $(H\setminus d)\cup d'$ is the unique block.
If $r=2$, write $Y\cap H=d$ and $Y\cap H'=C'$, where $|C'|=3$. There are only two possible types. A $(2,4)$ block exists exactly when the matching $\Theta(d)$ has a <duad> contained in $H'\setminus C'$. There is then exactly one such <duad>, since two disjoint <duads> cannot fit inside a triple. For a perfect matching on two triples, either all three pairs are cross, or there is one internal pair in each triple and one cross pair. Thus a $(2,4)$ block exists precisely when $\Theta(d)$ is not entirely cross for $C'\mid(H'\setminus C')$.
On the other hand, a $(3,3)$ block containing $Y$ must use the unique partition of $H$ corresponding to $C'\mid(H'\setminus C')$. It exists precisely when $d$ is contained in one of that partition's triples, and is then unique. By the partition incidence rule in part (b), this happens precisely when $\Theta(d)$ is entirely cross. Hence exactly one of the two possible block types exists, always uniquely.
For $r=3$, interchange the roles of $H$ and $H'$ and use the inverse partition incidence rule proved in part (b). More explicitly, the <duad> $Y\cap H'$ either has an inverse <syntheme> with an internal pair in the complement of the triple $Y\cap H$, yielding a unique $(4,2)$ block, or its inverse <syntheme> is entirely cross, yielding the unique $(3,3)$ block. These alternatives are exclusive and exhaustive for the same matching-on-two-triples reason.
Every split $r=0,\ldots,5$ therefore gives exactly one containing block. We have constructed
$$
\boxed{S(5,6,12),\text{ the small Witt design}.}
$$
As an independent count, each block contains six five-subsets and $132\cdot6=792=\binom{12}{5}$. The case proof establishes uniqueness and existence; the count alone would not have done so.
Back to article page
