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 $V$.

We start using only some divisibility conditions.

Theorem 1: 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 $p+1$.

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 $p+1$. $\Diamond$

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.

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

$$\vert S^G\vert \ = \ \frac{\vert G\vert}{\vert N_G(S)\vert}.$$

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.

We define

$$\lambda(T,K^G) \ = \ \vert\{K'\vert T\subset K'\in K^G\}\vert$$

and

$$\mu(T^G,K) \ = \ \vert\{T'\vert T'\subset K, T'\in T^G\}\vert.$$

Then we obtain a general formula already used by Alltop in 1965, [1].

Alltop's Lemma 2: 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

$$\vert T^G\vert \lambda(T,K^G) = \vert K^G\vert \mu(T^G,K)\vert.$$

This follows easily by doubly counting

$$\vert\{(S,B)\vert S \in T^G, B \in K^G, S\subset B\}\vert.$$

$\Diamond$

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

Corollary 3: Let $G$ act $t$-homogeneously in Alltop's Lemma. Then

$$\lambda(T,K^G) \ = \ \frac{{k\choose t}}{\vert N_G(K)\vert}.$$

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

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

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 [15].

Lemma 4 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 $g \in N_G(U)$.

Proof. We have

$$U^g = N_G(T_1)^g = N_G(T_1^g) = N_G(T_2) = U$$

such that $g \in N_G(U)$. $\Diamond$

If 5 divides $\vert PSL(2,p)\vert \ =\ \frac{1}{2}(p+1)p(p-1)$ then 5 divides $p-1$ by Theorem 1. We know from group theory [10] 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.

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

Theorem 5: 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 $\vert N_{PSL(2,p)}(T)\vert = 1.$ For the block $K$ containing $T$ of a $5$-$(p+1,6,1)$ design admitting $PSL(2,p)$ we have that $\vert N_{PSL(2,p)}(K)\vert$ divides $6$.

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, $\vert N_{PSL(2,p)}(T)\vert = 1.$ The stabilizer $N_{PSL(2,p)}(K)$ of a 6-set $K$ then has orbits of length $\vert N_{PSL(2,p)}(K)\vert$ on the 5-sets contained in $K$. So, $6 = \vert N_{PSL(2,p)}(K)\vert \times x$ where $x$ is the number of orbits. $\Diamond$

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 $\{1,2,3,5,6\}$.

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 $\frac{\vert G\vert}{i}$ of $G = PSL(2,p)$ on the blocks of the design. Then

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

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

$$b = {v\choose 5}/6.$$

Then,

$$\vert G\vert = \frac{v(v-1)(v-2)}{2}$$

has to be inserted into the equation.

After cancelling we obtain

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

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 $\vert PSL(2,p)\vert$ because also $a_5 \le 2$.

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.

Theorem 6 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 $\frac{\vert PSL(2,p)\vert}{i}$ 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 $p = 11$.

Proof. We only have to consider the cases $i \in \{1,2,3,5,6\}$. The case $i = 1$ is already contained in [5]. 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 $\frac{\vert PSL(2,p)\vert}{i}$ 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 $\frac{\vert PSL(2,p)\vert}{i}$. 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. $\Diamond$

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.