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Basic Conditions
In this section we investigate some necessary conditions that have to hold if
there exists a -
design with automorphism group
.
So we throughout assume
to be a point set of
points. A
-set
simply means a subset of size
of
.
We start using only some divisibility conditions.
Theorem 1:
If -
is an admissible parameter set for a
-design then
is
even. If in addition
for some prime
then
3 divides
and 5 does not divide
.
Proof
By double counting the number of blocks of a
-
design is always
![\begin{displaymath}
b = \lambda_t {v \choose t}/{k \choose t}
\end{displaymath}](img34.png)
where
![$\lambda_t = \lambda$](img35.png)
Because a -design is also a
-design with the same
number of blocks the corresponding formula for
yields
the
.
In case
and
then
such that
is even.
Continuing with in the same way then
such that
if
then 3 does not divide
and hence has to divide
The case
is clearly impossible.
Continuing in this way we obtain from the
formula for that 5 is a divisor of
such that 5 does not divide
.
We get some sharper results if we assume as a group
of automorphisms. Our main tool will be to analyse orbits of
on subsets
of the point set
by looking at the
set stabilizer
of such a subset.
We introduce some general notations. We assume that a finite group
acts on a finite set
of
points. For a subset
of
the length of the orbit
under
is
![\begin{displaymath}\vert S^G\vert \ = \ \frac{\vert G\vert}{\vert N_G(S)\vert}.\end{displaymath}](img55.png)
Orbits on -sets
and
-sets
for
are related by some numbers which are important for the
construction of
-designs.
We define
![\begin{displaymath}\lambda(T,K^G) \ = \ \vert\{K'\vert T\subset K'\in K^G\}\vert\end{displaymath}](img60.png)
and
![\begin{displaymath}\mu(T^G,K) \ = \ \vert\{T'\vert T'\subset K, T'\in T^G\}\vert.\end{displaymath}](img61.png)
Then we obtain a general formula already used by Alltop in 1965, [1].
Alltop's Lemma 2:
Let a group act on a set
and let
be a
-set and
a
-set of
. Then
![\begin{displaymath}\vert T^G\vert \lambda(T,K^G) = \vert K^G\vert \mu(T^G,K)\vert.\end{displaymath}](img64.png)
This follows easily by doubly counting
![\begin{displaymath}
\vert\{(S,B)\vert S \in T^G, B \in K^G, S\subset B\}\vert.
\end{displaymath}](img65.png)
![$\Diamond$](img50.png)
The group acts
-homogeneously if it is transitive on the
-sets.
Corollary 3:
Let act
-homogeneously in Alltop's Lemma. Then
![\begin{displaymath}\lambda(T,K^G) \ = \ \frac{{k\choose t}}{\vert N_G(K)\vert}.\end{displaymath}](img66.png)
Proof.
Substituting
,
, and
in the equation and cancelling
yields
the claimed result.
In the special case of and
we obtain that
divides 10.
In particular, if 5 does not divide
this implies
that
divides 2.
We will construct orbit representatives by constructing -sets invariant under a
prescribed subgroup
. 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 act on a set
and let
be two t-sets
having the same stabilizer
in
. Let
for some
.
Then
.
Proof.
We have
![\begin{displaymath}U^g = N_G(T_1)^g = N_G(T_1^g) = N_G(T_2) = U\end{displaymath}](img79.png)
such that
![$g \in N_G(U)$](img78.png)
![$\Diamond$](img50.png)
If 5 divides
then 5 divides
by Theorem 1. We know from group theory [10] that
then there exists only one conjugacy class of subgroups of order
in
. Such a subgroup is a dihedral group
with a cyclic normal subgroup of
order
. A subgroup
of order 5 of
then has orbits of length
5 and 1 only and
.
The
orbits of length 5 form one orbit of
on 5-element subsets.
Thus, by Lemma 4, there is only one orbit of
on 5-sets with a stabilizer of order 5.
Let us assume
to be
a
-
design admitting
. Then an orbit
of
of size 5 is contained in exactly one
block
of
. This block is of size 6 and has to be invariant under
.
So,
consists of
and an additional fixed point of
.
We have to choose one of the two fixed points of
to determine the first orbit
of
that has to belong to the design
. The two choices lead to isomorphic solutions,
since
maps one of the orbits onto the other one.
We now investigate other orbits of 5-sets and 6-sets.
Theorem 5:
Let
and
a 5-set whose stabilizer
in
has an order different from 5. Then
For the block
containing
of a
-
design admitting
we have that
divides
.
Proof.
A subgroup leaving a 5-set
invariant acts on
.
No elements of
different from the identity
has more than 2 fixed points. So, the orbits on
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
, no element of order 2 of
has any fixed point. So, (4,1), (2,2,1) do nor occur.
By Corollary 3, no element of order 3 fixes
. Therefore,
and
do not occur.
So,
The stabilizer
of a 6-set
then has orbits of length
on the 5-sets contained in
.
So,
where
is the number of orbits.
We conclude that all stabilizers of blocks of a Steiner
5-system -
with automorphism group
must have orders in
.
We want to derive some conditions on the numbers of blocks
which lie in orbits of a fixed length.
Let be the number of orbits of size
of
on the blocks of the design. Then
![\begin{displaymath}
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.
\end{displaymath}](img103.png)
From the general formula for the number of blocks in a -design we
obtain in this case
![\begin{displaymath}
b = {v\choose 5}/6.
\end{displaymath}](img104.png)
Then,
![](img105.png)
has to be inserted into the equation.
After cancelling we obtain
![](img106.png)
This equation can be reduced modulo some primes to get short relations.
Reducing modulo 5 gives our already known result that if
5 divides
because also
.
We consider possibilities of prescribing certain choices of stabilizers.
If only one of all is assumed to be non-zero then we obtain
the following results.
Theorem 6
Let a -
design with automorphism group
partition
into
-
designs with automorphism group
consisting of only one orbit of
of length
each.
Then either
and
or
and
or
and
.
Proof.
We only have to consider the cases
.
The case
is already contained in [5]. So let now
.
Then
![](img119.png)
We write this as
![](img120.png)
and reduce modulo 4,5, and 9. Since
![$v$](img31.png)
![$v$](img31.png)
![$v \equiv 3,\ 4 \ mod\ 5,\ 9$](img121.png)
![$p+1\equiv 4\ mod \ 180$](img122.png)
![$p+1\equiv 148\ mod \ 180$](img123.png)
![$p$](img12.png)
![$i = 2$](img114.png)
![$i=5$](img116.png)
![$5$](img2.png)
![$(p+1,6,1)$](img3.png)
![$5$](img2.png)
![$(12,6,1)$](img124.png)
![$\frac{\vert PSL(2,p)\vert}{i}$](img111.png)
![$5$](img2.png)
![$(p+1,6,1)$](img3.png)
![$i$](img125.png)
![\begin{displaymath}
b = {v\choose 5}/6
\end{displaymath}](img126.png)
of the 5-design has to be divided by
![$\frac{\vert PSL(2,p)\vert}{i}$](img111.png)
![$\frac{i}{60}(\frac{v}{3}-1)(\frac{v}{2}-2)$](img127.png)
![$\frac{v-12}{12}$](img128.png)
![$C_3$](img129.png)
![$\frac{v}{6} +1$](img130.png)
![$\Diamond$](img50.png)
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.
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Next: Subgroups of order up Up: Partitioned Steiner 5-Designs Previous: Introduction N.N. 2002-02-25