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\title{Partitioned Steiner 5-Designs }
\date{}
\author{Reinhard Laue, Alfred Wassermann,\\ University of Bayreuth, Germany}
\pagestyle{headings}
\begin{document}

\maketitle

\begin{abstract}
Orbits of $PSL(2,p)$ on 6-element subsets of the projective line with
prescribed non-trivial stabilizer are described. A refinement of cross-ratio
computations to $PSL(2,p)$ orbits allows to determine the orbits on
5-element subsets that they cover. Then Steiner $5$-$(p+1,6,1)$ designs 
are assembled from them. In particular, there is one isomorphism type
of $5$-$(48,6,1)$ designs that consists of $PSL(2,p)$ orbits of the same size,
each being a $3$-$(48,6,30)$ design. There are 7 isomorphism types
of $5$-$(84,6,1)$ designs of this type. Generally, Steiner $5$-$(p+1,6,1)$ designs
with such an orbit partition may only exist if $p\equiv 48,\ 84\ mod \ 180$.
 

\end{abstract}

\section{Introduction}
Up to now, there are only finitely many Steiner 5-designs known
and, with one exception of a 5$-$(36,6,1) design \cite{laue36},  they all
admit some $PSL(2,q)$ as a group of automorphisms. But $PGL(2,p)$ is not
admitted by a result of Denniston \cite{denniston80}. So, we here 
analyse the special situation
of $5$-$(p+1,6,1)$ designs with automorphism group $PSL(2,p)$
for some prime $p$. In addition, we only consider such designs
that are assembled from orbits with non-trivial stabilizers.
This restriction  has already been introduced by Denniston \cite{denniston80}
and was used by  Grannell, Griggs and Mathon 
\cite{GrannellGriggsMathon93} and
Betten, Laue, Molodtsov, Wassermann \cite{BLMW}. 
We show that these orbits of 6-element subsets can be constructed very fast by computer
and that one also can get the information on the orbits
of 5-element subsets  that are contained in a 6-element subset from a fixed
orbit. This is the information usually needed for a general backtrack solver
to collect orbits for a design. The approach is an obvious refinement of the
widely used methods of Kramer and Mesner \cite{krames} where all orbit
incidences are kept in a matrix, called the Kramer-Mesner matrix now.

\medskip
There are two techniques employed for this approach. Firstly, orbit calculations
are replaced by fixed point determination like in Polya's theory of counting.
We already made use of this idea in solving isomorphism problems for $t$-designs
with prescribed automorphism group \cite{goessweinstein}.
Secondly, the idea of using cross-ratio's as an invariant for the 5-orbits
employed by Alltop \cite{Alltop71} for 4-orbits, see also \cite{BJL99},
is refined from $PGL(2,p)$-orbits to $PSL(2,p)$-orbits. This variation
is worked out for the case of $p \equiv\ 3\ mod\ 4$, the only case where 
$5$-$(p+1,6,1)$ designs with automorphism group $PSL(2,p)$ may exist.
Here the Legendre symbol has to be evaluated for a certain ratio in addition
to the ordinary cross ratio.

\medskip
There are some divisibility conditions and group theoretic results that reduce the 
possibilities for $5$-$(p+1,6,1)$ designs with automorphism group $PSL(2,p)$ to exist.
The number $p+1$ of points has to be divisible by 12 and if 5 divides $p^2-1$ then
5 divides $p-1$. Also, the stabilizer of a block of a $5$-$(p+1,6,1)$ design in
$PSL(2,p)$ may only have an order $a \in  \{1,2,3,5,6\}$.
Since $PSL(2,p)$ is 3-homogeneous if  $p \equiv\ 3\ mod\ 4$, each orbit is a 3-design. 
If all orbits that form a $t$-design are of the same length we have a partition 
of the $5$-$(p+1,6,1)$ design into 3-designs all having the same parameters.
These might be useful for constructions like the famous Large Set recursions.
If we look for designs formed by orbits of the same length then we can use additional restrictions. 
We already considered the case of orbits of length $|PSL(2,p)|$ in \cite{BLMW}.
Then $p \equiv 83,\ 227\ mod\ 360$. We here show that the only further possibility
is that all orbits have length $\frac{|PSL(2,p)|}{2}$ and then $p \equiv\ 47,\ 83\ mod\ 180$.
We have found such designs for $p = 47$ and $p = 83$. Prescribing a stabilizer of
order 2 reduces the problem size drastically, but the next case $p = 228$ still
is too large for our presently used solver.

\medskip
The methods are implemented by an algorithm written in C. We make use of some
special shortcuts that only can be applied in a search for Steiner systems. In particular,
only those columns of the Kramer-Mesner matrix are kept which have only $0/1$-entries
and duplicate columns are avoided. The matrix is kept as a sparse matrix and the
invariant of a 5-element orbit is used for directly addressing the orbit.
No data structure is needed for the big group $PSL(2,p)$.
So, the computation time is linear in the number of orbits on 5-element sets and
6-element sets constructed.
The program allows to prescribe any combination of non-trivial stabilizers. We have 
collected some statistics in a table.
The first author thanks Gene Cooperman for inspiring discussions  and a fruitful time
at the Northeastern University that initiated this research. 



\section{Basic Conditions}

In this section we investigate some necessary conditions that have to hold if
there exists a $5$-$(p+1,6,1)$ design with automorphism group $PSL(2,p)$.
So we throughout assume $V$ to be a point set of $v = p+1$ points. A $n$-set
simply means a subset of size $n$ of $V$.

\medskip
We start using only some divisibility conditions.

\medskip
{\bf Theorem 1}:
{\em
If $5$-$(v,6,1)$ is an admissible parameter set for a $t$-design then $v$ is
even. If in addition
$v = p+1$ for some prime $p$ then 
3 divides $p+1$ and 5 does not divide $p+1$.
}

{\bf Proof}
By double counting the number $b$ of blocks of a $t$-$(v,k,\lambda)$
design is always 
$$
b = \lambda_t {v \choose t}/{k \choose t}
$$
where $\lambda_t =  \lambda$.

Because a $t$-design is also a $(t-1)$-design with the same
number of blocks the corresponding formula for $t-1$ yields
the $\lambda_{t-1}$.
In case $k = 6$ and $t = 5$ then $\lambda_{t-1} = (v-4)/2$
such that $v = p+1$ is even.

Continuing with $t-2$ in the same way then 
$\lambda_{t-2} = (v-3)(v-4)/6$ such that 
if $p \ne 3$ then 3 does not divide $v-4$
and hence has to divide $v = p+1.$
The case $p = 3$ is clearly impossible.

Continuing in this way we obtain from the
formula for $\lambda_1$ that 5 is a divisor of
$(v-1)(v-2)(v-3)(v-4)$ such that 5 does not divide $p+1$.
\hfill $\Diamond$\\


\medskip
We get some sharper results if we assume  $PSL(2,p)$ as a group
of automorphisms. Our main tool will be to analyse orbits of 
$PSL(2,p)$ on subsets $S$ of the point set $V$ by looking at the 
set stabilizer $N_{PSL(2,p)}(S)$ of such a subset.

\medskip
We introduce some general notations. We assume that a finite group $G$
acts on a finite set $V$ of $v$ points. For a subset $S$ of $V$
the length of the orbit $S^G$ under $G$ is
$$|S^G| \ = \ \frac{|G|}{|N_G(S)|}.$$

Orbits on $t$-sets $T$ and $k$-sets $K$ for 
$t \le k$ are related by some numbers which are important for the
construction of $t$-designs.

\medskip
We define 
$$\lambda(T,K^G) \ = \ |\{K'| T\subset K'\in K^G\}|$$
and
$$\mu(T^G,K) \ = \ |\{T'| T'\subset K, T'\in T^G\}|.$$ 
Then we obtain a general formula already used by Alltop in 1965, \cite{Alltop66}.


\medskip
{\bf Alltop's Lemma 2}:
{\em 
Let a group $G$ act on a set $V$ and let $T\subset V$
be a $t$-set and $K\subset V$ a $k$-set of $V$. Then 
$$|T^G| \lambda(T,K^G) = |K^G| \mu(T^G,K)|.$$
}

\medskip
This follows easily by doubly counting
$$
|\{(S,B)| S \in T^G, B \in K^G, S\subset B\}|.
$$ 
\hfill $\Diamond$\\

The group $G$ acts $t$-homogeneously if it is transitive on the $t$-sets.

{\bf Corollary 3}:
{\em
Let $G$ act $t$-homogeneously in Alltop's Lemma. Then 
$$\lambda(T,K^G) \ = \ \frac{{k\choose t}}{|N_G(K)|}.$$
}

\medskip
{\bf Proof}.
Substituting $\mu(T^G,K) = {k\choose t}$,  $|T^G| = \frac{|G|}{|N_G(T)|}$, and
$|K^G| = \frac{|G|}{|N_G(K)|}$ in the equation and cancelling $ |G|$ yields 
the claimed result.
\hfill $\Diamond$\\

\medskip
In the special case of $G = PGL(2,p)$ and
$t = 3,\ k = 5$ we obtain that $|N_{PGL(2,p)}(K)|$ divides 10.
In particular, if 5 does not divide $|PSL(2,p)|$ this implies
that $|N_{PGL(2,p)}(K)|$ divides 2.

\medskip
We will construct orbit representatives by constructing $k$-sets invariant under a
prescribed subgroup $U$. From these we first have to remove those invariant under a
larger group and then have to decide whether they lie in the same orbit.
This can be done using the next Lemma \cite{constrBurnside}.

\medskip
{\bf Lemma 4}
{\em
Let a group $G$ act on a set $V$ and let $T_1, \ T_2 \subset V$ be two t-sets
having the same stabilizer $U$ in $G$. Let $T_1 ^g = T_2$ for some $g \in G$.
Then $g \in N_G(U)$.
}

\medskip
{\bf Proof}.
We have $$U^g = N_G(T_1)^g = N_G(T_1^g) = N_G(T_2) = U$$
such that $g \in N_G(U)$.
\hfill $\Diamond$\\


\medskip
If 5 divides $|PSL(2,p)| \ =\ \frac{1}{2}(p+1)p(p-1)$
then 5 divides $p-1$ by Theorem 1. We know from group theory \cite{huppert} that
then there exists only one conjugacy class of subgroups of order $p-1$
in  $PSL(2,p)$. Such a subgroup is a dihedral group $D$ with a cyclic normal subgroup of 
order $\frac{p-1}{2}$. A subgroup $U$ of order 5 of $D$ then has orbits of length
5 and 1 only and $D = N_{PSL(2,p)}(U)$.
 The $\frac{p-1}{5}$ orbits of length 5 form one orbit of $D$ on 5-element subsets.
Thus, by Lemma 4, there is only one orbit of $PSL(2,p)$ on 5-sets with a stabilizer of order 5.
Let us assume $\cal D$ to be
 a $5$-$(p+1,6,1)$ design admitting
$PSL(2,p)$. Then an orbit $T$ of $U$ of size 5 is contained in exactly one
block $B$ of $\cal D$. This block is of size 6 and has to be invariant under $U$.
So, $B$ consists of $T$ and an additional fixed point of $U$.
We have to choose one of the two fixed points of $U$
to determine the first orbit $B^{PSL(2,p)}$ of $PSL(2,p)$
that has to belong to the design $\cal D$. The two choices lead to isomorphic solutions,
since $PGL(2,p)$ maps one of the orbits onto the other one.

\medskip
We now investigate other orbits of  5-sets and 6-sets.

\medskip

{\bf Theorem 5}:
{\em Let $ p \equiv 3 \ mod \ 4$ and $T$ a 5-set whose stabilizer $N_{PSL(2,p)}(T)$
in $PSL(2,p)$ has an order different from 5. Then $|N_{PSL(2,p)}(T)| = 1.$
For the block $K$ containing $T$ 
of a $5$-$(p+1,6,1)$ design admitting 
$PSL(2,p)$ we have that $|N_{PSL(2,p)}(K)|$ divides $6$.
}

\medskip
{\bf Proof}.
A subgroup $U \ne id$ leaving a 5-set $T$ invariant acts on $T$.
No elements of $PGL(2,p)$ different from the identity
has more than 2 fixed points. So, the orbits on $T$ must
be of type (5), (4,1), (3,2), (3,1,1), (2,2,1).
By assumption (5) does not occur.
Since 2 does not divide $(p-1)/2$, no element of order 2 of $PSL(2,p)$
has any fixed point. So, (4,1), (2,2,1) do nor occur.
By Corollary 3, no element of order 3 fixes $T$. Therefore,
$(3,2)$ and $(3,1,1)$ do not occur.
So, $|N_{PSL(2,p)}(T)| = 1.$
The stabilizer $N_{PSL(2,p)}(K)$ of a 6-set $K$ then has orbits of length
$|N_{PSL(2,p)}(K)|$ on the 5-sets contained in $K$.
So, $6 = |N_{PSL(2,p)}(K)| \times x$
where $x$ is the number of orbits.
\hfill $\Diamond$\\



We conclude that all stabilizers of blocks of a Steiner
5-system $5$-$(p+1,6,1)$ with automorphism group $PSL(2,p)$
must have orders in $\{1,2,3,5,6\}$.

\bigskip



\medskip
We want to derive some  conditions on the numbers of blocks
which lie in orbits of a fixed length.


Let $a_i$ be the number of orbits of size $\frac{|G|}{i}$
 of $ G = PSL(2,p)$ on the blocks of the design. Then

$$
b = a_1 |G| + a_2 |G|/2 + a_3 |G|/3 + a_5 |G|/5 + a_6 |G|/6.
$$


From the general formula for the number of blocks in a $t$-design we 
obtain in this case

$$
b = {v\choose 5}/6.
$$ 
Then,
$$
|G| = \frac{v(v-1)(v-2)}{2}
$$ has to be inserted into the equation.

After cancelling we obtain
$$
30 a_1 + 15 a_2 + 10 a_3 + 6 a_5 + 5 a_6 = (\frac{v}{3}-1)(\frac{v}{4}-1).
$$

\medskip
This equation can be reduced modulo some primes to get short relations.
Reducing modulo 5 gives our already known result that $a_5 = 1$ if
5 divides $ |PSL(2,p)|$ because also $a_5 \le 2$.

\medskip
We consider possibilities of prescribing certain choices of stabilizers.
If only one of all $a_i$ is assumed to be non-zero then we obtain
the following results.

\medskip
{\bf Theorem 6}
{\em
Let a $5$-$(p+1,6,1)$ design with automorphism group $PSL(2,p)$ partition
into $a$ $3$-$(p+1,6,\lambda)$ designs with automorphism group $PSL(2,p)$
consisting of only one orbit of $PSL(2,p)$ of length $\frac{|PSL(2,p)|}{i}$
each.
Then either $i = 1$ and $p+1 \equiv \ 84, \ 228 \ mod \ 360$ or
$i = 2$ and $p+1 equiv \ 48,\ 84 \ mod \ 180$ or $i=5$
and $p = 11$.
}

\medskip
{\bf Proof}.
We only have to consider the cases $i \in \{1,2,3,5,6\}$.
The case $i = 1$ is already contained in \cite{BLMW}. So let now $i = 2$.
Then 
$$ 
15  a_2 \ = \ (\frac{v}{3}-1)(\frac{v}{4}-1).
$$
We write this as 
$$12\cdot 15 a_2 \ = \ (v-3)(v-4)$$
and reduce modulo 4,5, and 9. Since $v$ is even, 4 divides $v$.
We have the solutions $v \equiv 3,\ 4 \ mod\ 5,\ 9$ such that we have four 
combinations modulo 180 by the Chinese Remainder Theorem. The solutions $p+1\equiv 4\ mod \ 180$
and $p+1\equiv 148\ mod \ 180$
would imply that 3 divides $p$. So, only the claimed cases remain for $i = 2$.
There is one further case for $i = 5$ where each of the only two existing orbits
of 6-sets with a stabilizer of order 5 is already a $5$-$(p+1,6,1)$ design. That is the well
known small Witt design with the parameters $5$-$(12,6,1)$.
In all other cases the number of orbits of length $\frac{|PSL(2,p)|}{i}$
needed to form a $5$-$(p+1,6,1)$ design is larger than the existing number of 
orbits on 6-sets with a stabilizer of order $i$. To get this number of required orbits
the number of bocks
$$
b = {v\choose 5}/6
$$ 
of the 5-design has to be divided by $\frac{|PSL(2,p)|}{i}$.
This number is $\frac{i}{60}(\frac{v}{3}-1)(\frac{v}{2}-2)$.
We will show in Theorem 7  that there exist only $\frac{v-12}{12}$ orbits with stabilizer $C_3$
and $\frac{v}{6} +1$ orbits with a stabilizer of order 6. In both cases there are
no solutions.
\hfill $\Diamond$\\

One can discuss further more complicated cases of selecting several stabilizers with
the same methods. We will give some experimental results on existing Steiner systems
in such situations below.






\newpage

\section{Subgroups of order up to 6 in $PSL(2,p)$}


The permutation group $PSL(2,p)$ is represented as the factor group 
$SL(2,p) Z/ Z)$, where $Z = Z(GL(2,p))$ consists of all 
non-zero multiples of the identity matrix. It is much easier to handle a
$2\times 2$-matrix than a permutation of length $p+1$, since we can apply
linear algebra to matrix problems. 
We therefore represent subgroups of $PSL(2,p)$ by the matrices which generate
them modulo $Z=Z(GL(2,p))$. 

\medskip
{\bf Theorem 7}
{\em
 Up to conjugation in $PSL(2,p)$ there are the following stabilizers and the
corresponding numbers of orbits of 6-sets. Here $a,b,c \in GF(p)$ are chosen
such that $a^2 + b^2 + 1 = ab$ and $c$ has order 5 if $5 | p-1$.
$$
\begin{array}{|l|c|c|}
\hline
  Group &  Generators\ mod\ Z & number\ of\ orbits \\
\hline
C_6   & 
 \begin{minipage}{6.5cm}\raisebox{-.01cm}{}\\
\[ \left( \begin{array}{cc}
2 & 1\\
-1 & 1
\end{array} \right) \] 
\raisebox{-.01cm}{}
 \end{minipage}
                              &  1 \\
\hline
S_3^{(1)}   & 
 \begin{minipage}{6.5cm}\raisebox{-.01cm}{}\\
\[ \left( \begin{array}{cc}
1 & 1\\
-1 & 0
\end{array} \right),
\left( \begin{array}{cc}
-a & b-a\\
b & a
\end{array} \right) \]     
\raisebox{-.01cm}{}
 \end{minipage}
                              &  \frac{p+1}{12} \\
\hline
S_3^{(2)}   & 
 \begin{minipage}{6.5cm}\raisebox{-.01cm}{}\\
\[ \left( \begin{array}{cc}
1 & 1\\
-1 & 0
\end{array} \right),
\left( \begin{array}{cc}
-b & a-b\\
a & b
\end{array} \right) \]     
\raisebox{-.01cm}{}
 \end{minipage}
                              &  \frac{p+1}{12} \\
\hline
C_3   & 
 \begin{minipage}{6.5cm}\raisebox{-.01cm}{}\\
\[ \left( \begin{array}{cc}
1 & 1\\
-1 & 0
\end{array} \right) \] 
\raisebox{-.01cm}{}
 \end{minipage}
                              &  \frac{p-11}{12} \\
\hline
C_2   & 
 \begin{minipage}{6.5cm}\raisebox{-.01cm}{}\\
\[ \left( \begin{array}{cc}
1 & 2\\
-2 & -1
\end{array} \right) \] 
\raisebox{-.01cm}{}
 \end{minipage}
                              &  \frac{(p+1)(p-9)}{24} \\
\hline
C_5   & 
 \begin{minipage}{6.5cm}\raisebox{-.01cm}{}\\
\[ \left( \begin{array}{cc}
1 & 0\\
0 & c
\end{array} \right) \] 
\raisebox{-.01cm}{}
 \end{minipage}
                              &  2 \ if\ 5| p-1 \\
\hline
\end{array}
$$
}


\medskip
{\bf Proof}

A subgroup $C_6$ of  $PSL(2,p)$ is generated by

\medskip
\[ \left( \begin{array}{cc}
2 & 1\\
-1 & 1
\end{array} \right) \] 
modulo $Z(GL(2,p))$.
It is easily veryfied that this matrix induces a permutation of order 6 on the
projective line. The determinant is 3 which is a quadratic residue mod $p$
exactly if 12 divides $(p-1)$ or $(p+1)$, see p. 92 of \cite{NivenZuckermann}.

There is just one orbit with this stabilizer on 6-sets.
A representative is the set $\{0,1,2,(p+1)/2,p-1,p\}$.

The 5-orbit covered by these 6-sets has as a stabilizer
in $PGL(2,p)$ the subgroup of order 2 generated by
\[ \left( \begin{array}{cc}
-1 & 0\\
1 & 1
\end{array} \right) \] modulo $Z(GL(2,p))$. 
This has the two fixed points $0,2$ and the 2-cycles
$(p,1),(i,i/(i-1))$ for $i \ne 1$. In particular,
$(p-1,(p+1)/2)$ is one such 2-cycle.


\medskip
The matrices 

\[ \left( \begin{array}{cc}
-1 & 0\\
1 & 1
\end{array} \right),
\left( \begin{array}{cc}
1 & 1\\
-1 & 0
\end{array} \right) \]      
modulo $Z(GL(2,p))$ generate an $S_3$ in $PGL(2,p)$.


\medskip
The matrices 

\[ \left( \begin{array}{cc}
1 & 1\\
-1 & 0
\end{array} \right),
\left( \begin{array}{cc}
-a & b-a\\
b & a
\end{array} \right) \]     
where $a^2 + b^2 + 1 = ab$, 
modulo $Z(GL(2,p))$ generate one $S_3$ in $PSL(2,p)$.
There are $(p+1)/12$ orbits with this stabilizer on 6-sets.

\medskip
By conjugating the matrix 
\[ \left( \begin{array}{cc}   
-a & b-a\\
b & a
\end{array} \right) \]  
with the matrix
\[ \left( \begin{array}{cc}
-1 & 0\\
1 & 1
\end{array} \right) \]
which is not in $PSL(2,p)$
we obtain another such subgroup:

\medskip
The matrices 


\[ \left( \begin{array}{cc}
1 & 1\\
-1 & 0
\end{array} \right),
\left( \begin{array}{cc}
-b & a-b\\
a & b
\end{array} \right) \]     
where $a^2 + b^2 + 1 = ab$, 
modulo $Z(GL(2,p))$ generate another $S_3$ in $PSL(2,p)$.
There are $(p+1)/12$ orbits with this stabilizer on 6-sets.

The last two subgroups isomorphic to $S_3$ are not conjugate in  $PSL(2,p)$. 
So, they are stabilizers of 6-sets from different orbits. The normalizer 
in $PSL(2,p)$ of each of the two subgroups is generated by $C_6$ and the chosen
subgroup. They have index 2 in their normalizer such that the 6-sets
invariant under one such $S_3$ fall into pairs that are mapped onto each other
under the action of the normalizer. By the homomorphism prinziple, these
pairs represent the orbits of $PSL(2,p)$ on  6-sets with stabilizer conjugate
to the chosen $S_3$.

All the listed elements modulo the center normalize the subgroup $C_6$ of $PSL(2,p)$
generated by
\[ \left( \begin{array}{cc}
2 & 1\\
-1 & 1
\end{array} \right). \]
It is easily veryfied that this matrix induces a permutation of order 6 on the
projective line. The determinant is 3 which is a quadratic residue mod $p$
exactly if 12 divides $(p-1)$ or $(p+1)$, see p. 92 of \cite{NivenZuckermann}.

\medskip
The subgroup $C_3$ of order 3 of $C_6$ is generated by
\[ \left( \begin{array}{cc}
1 & 1\\
-1 & 0
\end{array} \right) \] 
modulo $Z(GL(2,p))$.
This group has $c = (p+1)/3$ orbits on 3-sets. So, these yield
${c\choose 2}$ combinations for a 6-set invariant under $C_3$.

\medskip
We have to subtract those 6-sets first that are invariant under a 
subgroup of order 6 containing our $C_3$.
We have to subtract those $(p+1)/6$ which are invariant under $C_6$.
Each of the two $S_3$ has $(p+1)/12$ subgroups conjugate to it in the normalizer of 
our $C_3$, which is a dihedral group of order $p+1$.
So, we for these groups have to subtract 2 times  $(p+1)/12$ times  $(p+1)/6$
6-sets.
We are left with 
$$
{{(p+1)/3}\choose 2} - (p+1)/6 - 2(p+1)^2/(12\cdot 6)
$$
$$ = 
(p+1)/3 \cdot (p-11)/12
$$
6-sets which are invariant under $C_3$ but not
under any proper overgroup.
The orbits of $N_{PSL(2,p)}(C_3)$ on this set are the 
intersections of the $PSL(2,p)$-orbits with this set.
This normalizer has order $(p+1)$ and its orbits 
have length $(p+1)/3$, since $C_3$ is the stabilizer of each of theses 6-sets.
So, there remain $(p-11)/12$ orbits of 6-sets with 
$C_3$ as a stabilizer.




\medskip
The subgroup of order 2 of $C_6$ is generated by
\[ \left( \begin{array}{cc}
1 & 2\\
-2 & -1
\end{array} \right) \] 
modulo $Z(GL(2,p))$.

This element also represents the only conjugacy class 
of elements of order 2 in $PSL(2,p)$. Let us call the
subgroup it generates $U$. Then $U$ has $(p+1)/2$ orbits of
length 2. Any 6-set invariant under  $U$ is formed by selecting 
3 of these orbits. We apply Lemma 4 to determine the number of orbits 
of $PSL(2,p)$ on 6-sets where $U$ is a stabilizer of some representative.
We first have to subtract the 6-sets that are invariant under any
larger group. So, we have to subtract the $\frac{p+1}{6}$ 6-sets invariant under $C_6$.
Since all conjugates of $C_6$ have trivial intersection, our $C_2$ lies in only
one $C_6$.
There are 6-sets with a stabilizer of type $S_3$ and $C_2$ is contained in some
of these, because any subgroup of order 2 of an $S_3$ is conjugate to $U$.
As a group of order 6 each such $S_3$ has $\frac{p+1}{6}$ invariant  6-sets, namely its orbits. 
We now have to find out the number
$\lambda(U,S_3^{(i)PSL(2,p)})$ of conjugates of $S_3^{(i)}$ for $i = 1,2$ that contain $U$.
This can be done by doubly counting the set
$$\{(C,S)| C \in  U^{PSL(2,p)} \ and\ S \in S_3^{(i)PSL(2,p)} \ and\ C < S \}.$$
We obtain
$$
|U^{PSL(2,p)}|\lambda(U,S_3^{(i)PSL(2,p)}) = |S^{PSL(2,p)}|\cdot 3,
$$
and therefore
$$
\lambda(U,S_3^{(i)PSL(2,p)}) = 3\cdot \frac{|N_{PSL(2,p)}(U)|}{|N_{PSL(2,p)}(S_3)|}.
$$

The normalizer of $U$ is a dihedral group $D$ of order $(p+1)$ and 
the normalizer of $S_3^{PSL(2,p)}$ has order 12.
So, $\lambda(U,S_3^{(i)PSL(2,p)}) = \frac{p+1}{4}$ for each $i$.
By this we get the number of 6-sets with stabilizer $U$.
The number of orbits of $D$ on them is simply obtained by dividing by
the length of each orbit which is $\frac{|D|}{|U|} = \frac{p+1}{2}$.

So, we get
$$
\frac{2}{p+1}\{ {\frac{p+1}{2}\choose  {3}} - \frac{p+1}{6} - \frac{p+1}{2}\frac{p+1}{6}\}
= \frac{p+1}{6}\frac{p-9}{4}$$
orbits. 
\hfill $\Diamond$\\

\medskip
We illustrate some part of the subgroup lattice by the following picture.

\vspace{1cm}

\input{lattice.epic}

\medskip
While the counting of orbits is completely solved in this way the construction of orbit
representatives still needs some work. We have to find out whether a 6-set $K$
invariant under a subgroup $U$ has $U$ as its full set stabilizer. In the case of Steiner 
systems this can be done by the computation of 
$$ \lambda(T,K^{PSL(2,p)}) = \frac{|N_{PSL(2,p)}(T)|}{|U|}  \mu(T^{PSL(2,p)},K)$$
and discarding $K$ if this value is larger than 1.

\medskip
There are two possible reasons for such a value.
At first, $T$ may be contained in more than 1 block of the
orbit $K^{PSL(2,p)}$. Then this orbit cannot be in a Steiner system.
The second reason is that 
$U$ may not be the full
stabilizer of $K$. So, in a consideration of all representatives
of subgroups as possible stabilizers of some block a conjugate
of $N_{PSL(2,p)}(K)$ will occur stabilizing some block in $K^{PSL(2,p)}$.





\section {Invariants}
Each of the stabilizers of order 6 acts transitively on any invariant
6-set $K$. So, there is only one 5-set
orbit on $K$. Correspondingly, the column for $K^{PSL(2,p)}$
in the Kramer-Mesner matrix has only one entry 1 and all
other entries are 0. So, in these cases, we easily determine the entries of the
Kramer-Mesner matrix in these columns. The remaining cases are more complicated.
Here we use Alltop's Lemma. That requires for each 5-set $T$ contained in a representative
$K$ of a 6-set orbit with stabilizer $U$ to find out the 5-set orbit of $T$.
We do this by computing an invariant which separates the 5-set orbits.

\medskip
If a group acts $t$-transitive then any $t$-element sequence of pairwise different
elements can be mapped onto any other such sequence by some group element.
If the stabilizer of a sequence $S$ acts trivially then already the sequence of the 
image points of $S$ determines the image points of all other points.

To see this assume there would be different image points for some element $\omega$.
That would mean that there are two group elements $g,h$ such that 
$S^g = S^h$ but $\omega^g \ne \omega^h$. Then $gh^{-1}$ fixes each element of $S$
and thus should be trivial by assumption. But then $gh^{-1}$ also fixes $\omega$.

\medskip
Any sequence of pairwise different entries of length bigger than $t$
can be truncated after the first three entries. This is a projection
that is compatible with the group action.  So, two longer sequences
with identical starting sequence $S$ of length $t$ are in the same orbit
only if some element in the stabilizer of the first $t$ elements maps
one onto the other. If the stabilizer is trivial all these sequences 
lie in pairwise different orbits. This strategy of using mappings that are compatible with the group action
to simplify the problem is called {\em homomorphism principle} \cite{sogos}, \cite{goessweinstein}.
We have already used this approach in Lemma 4 where the homomorphism of group actions
is the assignment of the stabilizer to its k-set.


\medskip
If a group is regular on the orbit of $S$ then the orbit of each sequence $T$
with starting subsequence $S$  can be uniquely described by first applying the
unique group element $g$ that maps the starting sequence onto $S$ to $T$
and then using the sequence of the images of remaining points under $g$ as an
invarant.

\medskip
A prominent example of this kind is the cross-ratio of geometry.
The group $ GL(2,p^f)$ acts 3-transitive on the projective line
and regular on 3-sequences of pairwise different 1-dimensional subspaces.
By means of linear algebra a matrix can be found mapping one such sequence
onto a given second one. In particular, taking as the first sequence
a standard representative produces invariants.
For any 4-sequence the image of the fourth point after normalizing the first 3-element
sequence uniquely determines the orbit.

\medskip
In any subgroup  the stabilizer of three points acts 
trivially. So, if one first determines each orbit of the subgroup on the 3-sequences
and then uses the image of the fourth point one again has an invariant. We use this for
$PSL(2,p)$.
This group is 2-transitive and the stabilizer of two points is transitive on 
those points that lie in the same image of the determinant, that is those that are either
all squares or all non-squares.

\medskip
In this way we can map each sequence of 4 pairwise different points onto a unique
representative of its orbit. We define some ordering on the set of these representatives
and declare the smallest of all representatives from all sequences of
4 pairwise different points of a 5-set $K$ as the representative
of $K$.

\medskip
Of course this could be done as well for any $k$-sets. But for larger $k$ we get
too many 4-sets contained in a $k$-set. So, for $k = 6$ and our special goal
of finding Steiner systems we here make use of the following observation.
If a 6-set is contained in an orbit that already occured then the 5-orbits
it covers are already covered by the earlier found representative.
In terms of the Kramer-Mesner matrix that would mean that the column computed
for the new set is identical to an already existing one. So, we omit multiple
columns.

\medskip
There are cases where 6-sets from different orbits produce identical columns
of the Kramer-Mesner matrix. But if we look for Steiner systems we can use at
most one of two such columns. So, we could store for each column the multiplicity
of its occurrence and then multiply each solution with the factors of all columns
belonging to that solution. This would allow to count the complete number of solutions.
Then the number of isomorphism types is just half of the number of solutions
by \cite{goessweinstein}. This has not yet been implemented. 


 

\medskip
We work out the computation.
Take $<(1,0)>$ and $<(0,1)>$ as the representatives of the sequences of two 1-dimensional
subspaces. Then given any two subspaces $<(1,a)>$ and $<(1,b)>$ or $<(0,1)>$ we can determine a
matrix mapping the given sequence onto the sequence of representatives.

\medskip
The matrix 
\[ \left( \begin{array}{cc}
1 & -a\\
 0 & 1
\end{array} \right) \]                                                                                   
maps $(<(1,a)>,<(0,1)>)$ onto $(<(1,0)>,<(0,1)>)$ ,

\medskip
The matrix 
\[ \left( \begin{array}{cc}
a & 1\\
-1 & 0
\end{array} \right) \]                                                                                   
maps $(<(0,1)>,<(1,a)>)$ onto $(<(1,0)>,<(0,1)>)$ ,

\medskip
The matrix 
\[ \left( \begin{array}{cc}
b(b-a) & -a\\
  a-b& 1
\end{array} \right) \]                                                                                   
maps $(<(1,a)>,<(1,b)>)$ onto $(<(1,0)>,<(0,1)>)$ ,

\medskip
So, in each case we know the matrix $M$ transforming a given sequence of two
1-dimensional subspaces onto the standard basis. We have chosen the matrices so that 
the determinant $det(M)$ is a square modulo $p$.

\medskip
The matrix $M$ can be applied to all entries of a 4-sequence such that the first two
members are mapped onto  $<(1,0)>$ and $<(0,1)>$.
So, in a first step, the sequences are normalized in their first two components.
This can be done in constant time.

\medskip
We now consider the second step, where the third component is normalized.
The first two components now have to remain unchanged.
The subgroup of all matrices fixing the two subspaces  $<(1,0)>$ and $<(0,1)>$ 
consists of the matrices
of the form $diag(1,z)$ where $z \ne 0$. There are $p-1$ such matrices. 
Since any matrix that fixes 3 1-dimensional subspaces must fix all 1-dimensional subspaces,
this subgroup acts regularly on the set of subspaces of the form  $<(1,a)>$ for $a = 1,\cdots ,p-1$.
We chose as a representative $<(1,1)>$. The  matrix $diag(1,a^{-1})$ normalizes the third component.
So, there is only one orbit on 3-sequences under
$PGL(2,p)$.

\medskip
A sequence $$(<(1,a)>,<(1,b)>,<(1,c)>,<(1,d)>)$$
is transformed into normal form by first transforming by
\[ \left( \begin{array}{cc}
b(b-a) & -a\\
  a-b& 1
\end{array} \right) \]  
and then by 
\[ \left( \begin{array}{cc}
1 & 0\\
0 & \frac{(a-b)(c-b)}{c-a}
\end{array} \right) \]  
So, the product of these transformations 
\[ \left( \begin{array}{cc}
-b(a-b) & -a\frac{(a-b)(c-b)}{c-a}\\
a-b & \frac{(a-b)(c-b)}{c-a}
\end{array} \right) \]  
sends $<(1,d)>$ to $<(1,\frac{(a-d)(b-c)}{(b-d)(a-c)})>$.

\medskip
The quotient $$\frac{(d-a)(c-b)}{(d-b)(c-a)}$$ thus is an invariant of the orbit
of the given sequence under $PGL(2,p)$. This  quotient is known as the
{\em cross-ratio} in finite geometry. 

\medskip
In case of the special linear group the diagonal matrix has to lie in the coset of a matrix with determinant 1
modulo the kernel of the action of $PGL(2,p)$ on the set of all 1-dimensional subspaces.
This kernel consists of all matrices of the form 
 \[ diag(y,y) \ = \ \left( \begin{array}{cc}
y & 0\\
0 & y
\end{array} \right) \]  
 such that
their determinant is a square.
The squares form a subgroup of index two in the multiplicative group
of residues modulo $p$, which therefore has $\frac{p-1}{2}$ elements.
So, the $p-1$ subspaces lie in two orbits of length $\frac{p-1}{2}$ 
each. A representative of the first orbit is $<(1,1)>$ and the orbit
consists of all $<(1,z)>$ where $z$ is  a square. 
If 4 divides $p+1$ then 4 does not divide $p-1$. So,
the multiplicative group of residues modulo $p$ has no element
of order 4 such that there is no $y$ with $y^2 = -1$.
We have found the non-square $-1$ which gives rise to a representative
$<(1,-1)>$ of the other orbit. 

\medskip
We now have seen  that there are two orbits of PSL(2,p) on the set of 
sequences of length 3 with representatives
$$(<(1,0)>,<(0,1)>,<(1,1)>)$$
and
$$(<(1,0)>,<(0,1)>,<(1,-1)>).$$
If we apply the matrix 
\[ \left( \begin{array}{cc}
0 & -1\\
1 & 0
\end{array} \right) \]  
of determinant 1 to the first sequence we obtain a permuted version
of the second representatives. This shows that the corresponding
3-element sets of 1-dimensional subspaces are in one orbit. This means that
$PSL(2,p)$is {\em 3-homogeneous}.


\medskip
For our goal of normalizing a given sequence of 1-dimensional subspaces
we, after having transformed the sequence such that the first two
components are normalized,  have to multiply with some 
 $<(1,z)>$ where $z$ is  a square such that we either get 
$<(1,1)>$ or 
$<(1,-1)>$ as the third component.
The first case is just the cross-ratio derived above.
The second one is obtained similarly. It turns out that then 
 $<(1,d)>$ is sent to $<(1,\frac{(d-a)(c-b)}{(d-b)(a-c)})>$.
So, in this case we have to multiply the cross-ratio by -1.
That allows to use the well known mathematical term of a Legendre symbol.
For $p$ a prime and $0<a<p$ one has 
$$
\left( \frac{a}{p}\right) \ = \ 1$$
if $a$ is a quadratic residue modulo $p$ and 
$$
\left( \frac{a}{p}\right) \ = \ -1$$
if $a$ is a non-quadratic residue modulo $p$.
So, if $4|p+1$ then the invariant is the product of the cross-ratio
and the Legendre symbol.

\medskip
Using this invariant, the orbit of a 4-sequence can be 
determined in constant time.

\medskip
Now we have achieved a sequence of the form
$$
(<(1,0)>,<(0,1)>,<(1,1)>,Y)
$$
or 
$$
(<(1,0)>,<(0,1)>,<(1,-1)>,Y)
$$
where $Y = <(1,y)>$ or $Y = <(0,1)>$.
By the homomorphism principle, applied to the projection to the first 3 
components, two such sequences could be in the same orbit only
if they are mapped one onto the other by some element in the 
stabilizer of the firest three components. This stabilizer is
trivial such that all choices of $Y$ in the two sequences belong to
pairwise different orbits. The pair $<(1,1)>,Y$
or $<(1,-1)>,Y$  thus is an 
invariant for the orbits on sequences of length 4.


Of course, this can be continued to sequences of length 5 and longer.


\medskip
We now consider 4-element subsets instead of sequences.
They can be obtained from the sequences by the fusion version
of the homomorphism principle. 
A given 4-set $T$ can be arranged as a sequence 24 times.
So, we will get 24 values 
of our invariant for $T$.
Each transformation also transforms $T$ to some $T'$.
The sequences we can form from $T'$ are in the same orbits
as the sequences we have formed from $T$.
So, we don't have to iterate this process.
We can define the set $T'$ coming from the lexicographically smallest 
sequence among the 24 sequences as the canonical representative.
We also can use the transformation which transformed the
selected  sequence
into canonical form for transforming $T$ into canonical form.
In fact, if we take the 24 sequences of pairwise different points
from $T'$ then by normalizing them we will get back those that we obtained from $T$.
This shows that there is a bijection between the set of all 24 values of the
invariant and the selected smallest one.

\medskip
Alltogether we have presented a constant time algorithm which computes
for any given 4-element subset the canonical representative of its orbit
and a matrix which does this transformation.

A slight variation should be made when many 4-sets have to be transformed
into canonical form. Then searching each time for a square $z$ 
such that 
$$<(1,c) 
 \left( \begin{array}{cc}
1 & 0\\
0 & z
\end{array} \right) > \
=  \ <(1,x)>
$$
where $x = 1$ or $x=-1$
would take too much time. So, a table can be computed first which contains
for each value of $c$ the corresponding square $z$. This can be done in linear time, 
since either $z$ is the inverse of $c$ and then $c$ is a square
or $-z$ is the inverse of $c$ and then $c$ is a non-square.
So, besides a table of inverses another table of squares is needed.

\medskip
The stabilizer of $T'$ is obtained by taking the transforming elements that
map one sequence onto another one that also lies in $T'$.
The order of the stabilizer is obtained by just the number of transformed sequences
that lie in $T'$.

\medskip
From the orbits on 4-sets we can in a further step get the orbits on 5-sets.
If a 5-set $K$ is given, then one can form 5 partitions $K = T \cup \{x\}$
into a sequence of a 4-set $T$ and the remaining element. Transform $K$ by
the same element of $PSL(2,p)$ that maps $T$ onto its canonical representative $T'$.
Then the stabilizer of $T'$ may map the transformed point $x'$ onto a smaller one.
Choose the smallest one to get a representative on the set of all partitions
of a 5-set into two components of sizes 4 and 1.

\medskip
The stabilizer will be trivial in the case where we know in advance that already the
stabilizers of all 5-sets are trivial. So, then we need not compute them in this case.
We will get 5 different orbits of partitions that fuse to the same 5-set. The smallest
of them defines the canonical representative for our 5-set.


\medskip
The discussion yields the following result.

\medskip
{\bf Definition}
For a sequence $S$ of 4 pairwise different 1-dimensional subspaces of $V(2,p)$
let the {\em cross-ratio}  $cr(S)$ of $S$ and the {\em square indicator} $si(S)$ of $S$
be defined by the following table.

\bigskip
\begin{tabular}{|c|c|l|}\hline
$S$ & $cr(S)$ & $si(S)$\\
\hline
$(<(1,a)>,<(1,b)>,<(1,c)>,<(1,d)>)$ & $\frac{(a-d)(b-c)}{(a-c)(b-d)}$ &
 \begin{minipage}{6.5cm}\raisebox{-.01cm}{}\\
1 if $\frac{a-c}{(a-b)(b-c)}$ is a quadratic residue\\
-1 else\\
\raisebox{-.01cm}{}
 \end{minipage}\\
\hline
$(<(1,a)>,<(1,b)>,<(1,c)>,<(0,1)>)$ & $\frac{(b-c)}{(a-c)}$ &
 \begin{minipage}{6.5cm}\raisebox{-.01cm}{}\\
1 if $\frac{a-c}{(a-b)(b-c)}$ is a quadratic residue\\
-1 else\\
\raisebox{-.01cm}{}
 \end{minipage}\\
\hline
$(<(1,a)>,<(1,b)>,<(0,1)>,<(1,d)>)$ & $\frac{(a-d)}{(b-d)}$ &
 \begin{minipage}{6.5cm}\raisebox{-.01cm}{}\\
1 if $a-b$ is a quadratic residue\\
-1 else\\
\raisebox{-.01cm}{}
 \end{minipage}\\
\hline
$(<(1,a)>,<(0,1)>,<(1,c)>,<(1,d)>)$ & $\frac{(a-d)}{(a-c)}$ &
 \begin{minipage}{6.5cm}\raisebox{-.01cm}{}\\
1 if $c-a$ is a quadratic residue\\
-1 else\\
\raisebox{-.01cm}{}
 \end{minipage}\\
\hline
$(<(0,1)>,<(1,b)>,<(1,c)>,<(1,d)>)$ & $\frac{(b-c)}{(b-d)}$ &
 \begin{minipage}{6.5cm}\raisebox{-.01cm}{}\\
1 if $b-c$ is a quadratic residue\\
-1 else\\
\raisebox{-.01cm}{}
 \end{minipage}\\
\hline
\end{tabular}

\samepage{
\bigskip
{\bf Theorem}
{\em
Let $S$ be a sequence of 4 pairwise different 1-dimensional subspaces
of $V(2,p)$.
Then the orbit of $S$ under $GL(2,p)$ is uniquely determined by $cr(S)$.
The sequence $S$ is transformed into normal form
$$(<(1,0)>,<(0,1)>,<(1,1)>,<(1,cr(S))>)$$
by the matrix 
$$
A \ = \ \left\{ \begin{minipage}{16cm}
\begin{minipage}{6cm}
\[ \left( \begin{array}{cc}
-b(a-b) & -a\frac{(a-b)(c-b)}{c-a}\\
a-b & \frac{(a-b)(c-b)}{c-a}
\end{array} \right) \]
\end{minipage}
if the first three subspaces are $\ne \ <(0,1)>$\\

\begin{minipage}{6cm}
\[ \left( \begin{array}{cc}
-b(a-b) & -a(a-b)\\
a-b & a-b
\end{array} \right) \]  
\end{minipage}
if the third subspace  is $<(0,1)>$\\

\begin{minipage}{6cm}
\[ \left( \begin{array}{cc}
1 & \frac{a}{a-c}\\
0 & \frac{1}{c-a}
\end{array} \right) \] 
\end{minipage}
if the second subspace  is $<(0,1)>$\\

\begin{minipage}{6cm}
\[ \left( \begin{array}{cc}
b & b-c\\
-1 & 0
\end{array} \right) \] 
\end{minipage}
if the first subspace  is $<(0,1)>$\\
\end{minipage}
\right.
$$

If 4 divides $p+1$ then the orbit of $S$ under $SL(2,p)$ is uniquely determined by $cr(S)$
and $si(S)$.
The sequence $S$ is transformed into normal form
$$(<(1,0)>,<(0,1)>,<(1,si(S))>,<(1,si(S)\cdot cr(S))>)$$
by the matrix
$$
A \ = \ \left\{ \begin{minipage}{16cm}
\begin{minipage}{7cm}
\[ \left( \begin{array}{cc}
-b(a-b) & si(S)\cdot (-a)\frac{(a-b)(c-b)}{c-a}\\
a-b &si(S)\cdot  \frac{(a-b)(c-b)}{c-a}
\end{array} \right) \]
\end{minipage}
if the first three subspaces are $\ne \ <(0,1)>$\\

\begin{minipage}{7cm}
\[ \left( \begin{array}{cc}
-b(a-b) &si(S)\cdot ( -a)(a-b)\\
a-b &si(S)\cdot ( a-b)
\end{array} \right) \]
\end{minipage}
if the third subspace  is $<(0,1)>$\\

\begin{minipage}{7cm}
\[ \left( \begin{array}{cc}
1 &si(S)\cdot  \frac{a}{a-c}\\
0 &si(S)\cdot  \frac{1}{c-a}
\end{array} \right) \]
\end{minipage}
if the second subspace  is $<(0,1)>$\\

\begin{minipage}{7cm}
\[ \left( \begin{array}{cc}
b &si(S)\cdot ( b-c)\\
-1 &  0
\end{array} \right) \]
\end{minipage}
if the first subspace  is $<(0,1)>$\\
\end{minipage}
\right.
$$

}

\medskip

Using these formulae the time complexity of determining the orbit of a 4-element
sequence of pairwise different 1-dimensional subspaces is dominated by forming
a constant number of arithmetic operations and deciding whether some number
is a quadratic residue modulo $p$.
The last problem can be solved in time $O(log(p))$ using the Gau\ss \ reciprocity
formula, \cite{wolfart}, by a strategy similar to Euclid's gcd algorithm.

For a small prime $p$ and many tasks of computing canonical forms it is a better
strategy to first set up tables of squares and of inverses in linear time and then
for each canonical form computation using only constant time.


$$ 
\begin{array}{|r|c|c|c|c|c|c|l|l|}
\hline
\multicolumn{9}{|c|}{\rule[-3mm]{0mm}{8mm}\bf $5$-$(p+1,6,1)$\ Designs\ with\ Automorphism\ Group\ PSL(2,p)}\\
\hline
  $p$ &  C6 & C3 & S3 & C2 & C5 & Id &  no.\ of\ solutions/isom.-types & citation\\
\hline
\hline
  11  &  -  &  -  &  - &   -  &  x  & - &  1\  isom.\  type  &   \cite{ Witt38}\\
\hline
  23  &  x  &  -  &  x &   x  &     & - &    3\  isomorphism\  types & \cite{Denniston76} \\
  23  &  -  &  -  &  x &   x  &     & - &    2\  isomorphism\  types & \\
  23  &  x  &  -  &  - &   x  &     & - &    1\  isomorphism\  type & \\
\hline
  47  &  - &  - &  - &  x  &  & - &  {\bf 1\ isomorphism\ type }& \\
  47  &  - &  x &  - &  x  &  & - &   8\  solutions &\\
  47  &  x &  x &  - &  x  &  & - &  12\  solutions &\\
  47  &  - &  x &  x &  x  &  & - &  80\  solutions &\\
  47  &  x &  x &  x &  -  &  & - &   0\  solutions &\\
  47  &  - &  - &  x &  x  &  & - &  32\  solutions &\\
  47  &  x &  x &  x &  x  &  & x & 459\  isomorphism\  types  &  \cite{Denniston76}, \cite{GrannellGriggsMathon93}\\
\hline
  71 &   - &  - &  - &  x &  x & -  &      0\  solutions & \\
  71 &   - &  - &  x &  x &  x & -  &    262\  solutions & \\
  71 &   - &  x &  - &  x &  x & -  &      6\  solutions & \\
  71 &   x &  - &  - &  x &  x & -  &      0\  solutions & \\
  71 &   x &  x &  - &  x &  x & -  &     17\  solutions & \\
  71 &   x &  - &  x &  x &  x & -  &    696\  solutions & \\
  71 &   x &  x &  x &  - &  x & -  &      0\  solutions & \\
  71 &   - &  x &  x &  x &  x & -  &    352\  solutions & \\
  71 &   x &  x &  x &  x &  x & -  &   1501\  solutions/ 4204\ isomorphism\ types & \cite{Mills78}, \cite{GrannellGriggsMathon98}\\
  71 &   x &  x &  x &  x &  x & x  &   916299\  isomorphism\  types & \cite{BLMW}\\
\hline
  83 &   - &  - &  - &  x &    & -  &  {\bf   7\  isomorphism\  types} & \\
  83 &   x &  - &  - &  x &    & -  &     14\  solutions & \\
  83 &   - &  x &  - &  x &    & -  &     26\  solutions & \\
  83 &   - &  - &  x &  x &    & -  &   7440\  solutions & \\
  83 &   x &  x &  - &  x &    & -  &     36\  solutions & \\
  83 &   x &  x &  x &  - &    & -  &      0\  solutions & \\
  83 &   - &  x &  x &  x &    & -  &  29666\  solutions & \\
  83 &   x  & - &  x &  x &    & -  &  12820\  solutions & \\
  83 &   x &  x &  x &  x &    & -  &  51846\  solutions,\  38717\  isomorphism\  types & \cite{Denniston76},\cite{GrannellGriggsMathon98}\\
  83 &   - &  - &  - &  - &    & x  &  {\bf  3\  isomorphism\  types}  &\cite{BLMW}\\
\hline
 107 &   x &   - &   - &  x &    & -  &      0 & \\
 107 &   - &   - &   x &  x &    & -  &   >=\ 100000\ isomorphism\ types & \\
 107 &   x &   x &   x &  - &    & -  &      0 & \\
 107 &   - &   x &   - &  x &    & -  &   1136\ solutions & \\
 107 &   x &   x &   x &  x &    & -  &   >= 1000\ isomorphism\ types & \cite{GrannellGriggsMathon93}\\
\hline
 131 &   - &  - &  -  &   x &  x & -  &  ? & \\
 131 &   - &  - &  x  &   x &  x & -  & & \\
 131 &   x &  x &  x  &   - &  x & -  &      0 & \\
 131 &   x &  x &  x  &   x &  x & -  &    >= 1839\ running /X &\cite{GrannellGriggsMathon93}\\
\hline
 167 &   - &  - &  x  &   x &    & -  &    ?    & \\
 167 &   x &  x &  x  &   - &    & -  &    0  & \\
 167 &   x &  x &  x  &   x &    & -  &   >=4 \ isomorphism\ types & \cite{mathon}\\
\hline
\end{array}
$$

\newpage
\medskip
{\bf Remarks}

The next open case is $PSL(2,191)$.

\medskip
For prime powers there are further results. 
There exist at least $303$ isomorphism types of $5$-$(244,6,1)$ designs 
with automorphism group $P\Sigma L(2,3^5)$ \cite{BLMW}.

\medskip
Besides that there exists one $5$-$(36,6,1)$ design with automorphism group
$C_2 \times PGL(2,17)$.

\medskip
Most remarkable are the partitionable Steiner systems consisting of orbits
of the same size.
We have obtained exactly 1 isomorphism type for a prescribed stabilizer $C_2$
and $p = 47$ and exactly 7 isomorphism types 
for $p =83$.

\medskip
The next case would be
{\bf v = 228:}

\medskip
We would need 280 orbits with stabilizer $C_2$ or 140 orbits with trivial stabilizer
for a $5$-$(228,6,1)$ design.
So, we need only half as many orbits if we prescribe a trivial stabilizer 
as we need with stabilizer $C_2$.
But there are by far too many orbits with trivial stabilizer to choose the 140 orbits from.
So, it might be easier to prescribe $C_2$.

\medskip
The total number of $PSL(2,227)$ orbits on 6-sets is 32300.
Out of these only 2070 have length $\frac{|PSL(2,p)|}{2}$.
The number of 5-orbits is 840. In the Kramer-Mesner Matrix, reduced to these
6-orbits, in each column there are exactly 3 entries 1 and all others are 0.

\medskip
The effect of the  space reduction by using only a sparse matrix can be seen with $
PSL(2,443)$ where prescribing stabilizer $C_2$ results in a matrix of 
3234 5-orbits by 8028 6-orbits. The full matrix requires 52   MB
while the sparse matrix only needs 0.53 MB.
This problem size is far out of our present reach.

\medskip 
%Prescribing stabilizers $C_2$ and $S_3$ in $PSL(2,167)$ yields a Kramer-Mesner matrix of size 451x1128.


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\end{document}
