Steiner Triple Systems block clique Given a Steiner Triple System (STS) of order $v$, one can build its graph in the following way: each vertex is a block, and two verticies are adjacent if their blocks have nonempty intersection.
Thr proof of lemma 2.4 in The Steiner Triple Systems of Order 19 states that for $v \geq 19$ "[a] short case-by-case analysis shows that a set of blocks with pairwise nonempty intersection in an $STS(v)$ has size at most $7$ unless the blocks share a common point."
Can someone explain why this is the case?
Thanks.
 A: Let ${\cal F}=\lbrace B_k\rbrace_{1\leq k \leq n}$ be a set of blocks with pairwise nonempty intersection and $n\geq 7$. Note that $|B_i\cap B_j| \leq 1$ for every $i\neq j$
by definition of a Steiner system. We deduce 
$$|B_i\cap B_j|=1 \ (i < j) \tag{1}$$ 
If any four elements (call them $E_1,E_2,E_3,E_4$) of $\cal B$ 
share a common element $x$, then any other element (call it $F$) 
of $\cal B$ must share  an element $e_i$ with $E_i$ for $1\leq i\leq 4$. If the $e_i$ 
were all different from $x$, they would be pairwise distinct and this
would force $|F|\geq 4$ which is excluded. So at least one of the
$e_i$ is $x$, and hence $x$ is a common element to all the members
of $\cal F$ : we are thus done in this case. We can therefore assume
$$B_i\cap B_j\cap B_k \cap B_l=\emptyset \ (i<j<k<l) \tag{2}$$
Next, suppose there are  three elements (call them $E_1,E_2,E_3$) 
of $\cal B$ share a common element $x$. Then there are six distinct 
elements $a_i,b_i(1\leq i \leq 3)$ such that 
$E_i=\lbrace x,a_i,b_i \rbrace $. Let ${\cal G}=
{\cal F} \setminus \lbrace E_1,E_2,E_3\rbrace$. By (2), no member
of $\cal G$ contains $x$. By (1), we see that any $G\in{\cal G}$
shares an element $e_i\in \lbrace a_i,b_i \rbrace$ with $E_i$
for $1\leq i \leq 3$. The $e_i$ are all distinct, so 
$G=\lbrace e_1,e_2,e_3 \rbrace$ and hence 
${\cal G} \subseteq {\cal H}$ where 
${\cal H}=\lbrace a_1,b_1 \rbrace \times \lbrace a_2,b_2 \rbrace
\times \lbrace a_3,b_3 \rbrace$. Denote by $\tau_i$ the involution
interchanging $a_i$ and $b_i$, and let $\tau=(\tau_1,\tau_2,\tau_3)$. Then
$\tau$ is an involution of $\cal H$, and for any $H\in {\cal H}$, 
$\cal G$ cannot contain both $H$ and $\tau(H)$. So $\cal G$ contains 
at most half the elements of $\cal H$, so $|{\cal G}|\leq 4, |{\cal F}|\leq 7$.
we are thus done in this case. We can therefore assume
$$B_i\cap B_j\cap B_k =\emptyset \ (i<j<k) \tag{3}$$
When (3) holds, there are five distinct elements $a_i(1\leq i\leq 3)$
such that $B_1=\lbrace a_1,a_2,a_3\rbrace,B_2=\lbrace a_1,a_4,a_5\rbrace$.
By (3) again we have $a_1\not\in B_3$, and renaming if necessary
we can assume that there are yet another distinct element $a_6$ such
that $B_3=\lbrace a_2,a_4,a_6\rbrace$. Let ${\cal G}={\cal F} \setminus \lbrace B_1,B_2,B_3\rbrace$. By (3), no member $G$ of $\cal G$ can contain 
any appearing twice already in $B_1,B_2,B_3$, so $G$ cannot contain
$a_1,a_2$ or $a_4$. Using (1), we deduce $G=\lbrace a_3,a_5,a_6 \rbrace$,
so $|{\cal G}|\leq 1, |{\cal F}|\leq 4$ and this finishes the proof.
