BULL. AUSTRAL. MATH. SOC.
MOS O5C2O, 20B25
VOL. 6 (1972), 53-59.
An algebraic characterization of
finite symmetric tournaments
J.L. Berggren
A tournament
T
is called symmetric if its automorphism group
is transitive on the points and arcs of
this paper is that if
T
T
1 .
is a finite symmetric tournament then
is isomorphic to one of the quadratic residue tournaments
GF (pn) , pn = 3 (h) , by
formed on the points of a finite field
the following rule:
directed from
a
If
to
a, b £ GF(p )
b
quadratic residue in
exactly when
then there is an arc
b - a
Throughout this paper
\T\ . We call
T
T
will denote a finite tournament,
a symmetric tournament if
points and arcs of
T .
is directed from
a
is a non-zero
GF(p ) .
automorphism group, and the number of points in
T
The main result of
If
to
a
and
b
b
T
A(T)
its
is transitive on the
are points of
we write
A(T)
will be written as
T
and if an arc of
a > b .
The purpose of this paper is to prove the following theorem.
THEOREM
finite field
A.
Let
T
be a symmetric tournament.
GF(pn) , where
pn = 3 I1*) , such that
the tournament formed on the points of
a, b i GF(p )
x i GF(p } .
then
a > b
if
b - a = x
Moreover, identifying
group of all permutations of
T
GF(pn)
T
Then there is a
T
is isomorphic to
by the following rule:
If
, for some non-zero
with this tournament,
of the form
A(r)
is the
a -*• x a (a) + c , where
c
Received 18 August 1971.
53
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54
J.L. Berggren
GF(pn} ,
ranges over all elements of
GF(p")
3
and
o
REMARK.
x
over all non-zero elements of
over all field automorphisms of
Assuming that
T
GF (pn) .
is a tournament formed from
GF(p")
in the
manner described above, Goldberg [2] proved that the automorphism group of
T
is as described in the last sentence of Theorem A.
However, this result
is an immediate corollary of our classification of symmetric tournaments,
so we also include it here.
We first state and prove a lemma.
The various parts of this lemma are
well-known, but we include the statement and proof for completeness.
LEMMA
1.
Let
T
be a symmetric tournament.
The following
statements are true:
(i)
A(T)
has odd order and (so) is solvable;
(ii)
A(T)
is a primitive permutation group on the points of
(Hi)
there is a prime
p = 3 C1*) and an odd integer
n
T ;
such that
\T\ = pn .
Proof.
A(T)
Since no element of
A(T)
can interchange two points of
T ,
has odd order and is therefore solvable [J]. To prove (ii) suppose
B , ..., B
is a system of imprimitivity for
\B \ > 2 .
a > a
Let
a, b € B
and
a € B
and
ir(i>) = o .
Thus
TT(B )
r > 2
. We may assume
then by arc transitivity there exists
ir(a) = a
A(T) , so
TT f A(T)
intersects both
and
a > b .
If
such that
B
and
B
non-trivially, which is a contradiction of the definition of a system of
imprimitivity.
The case
acts primitively on
T .
c > a
also yields a contradiction.
Finally, we prove (iii).
conclude from Satz 3.2(a) [3, Chapter II] that
p .
For
a t. T
on the points of
let
T ,
0(a) = {b € T \ a > b} .
\0{a)\ = \0(b)\
for all
|0(a)| = (p"-l)/2 , so by arc-transitivity
A(T)
, the stabilizer of
p = 3 CO
and
n
a
in
A(T) .
A(T)
By (i) and (ii) we may
| T | = pK
Since
for some prime
A(r)
a, b € T .
(p"-l)/2
Hence
Hence
is transitive
This implies
divides the order of
(pM-l}/2
is odd and so
is odd.
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Finite symmetric tournaments
Before proceeding we state some notation.
n > 0
let
of order
V = V(n, p)
p
and let
be an
For a prime
and a £ 7
following rule:
I
by
If
and integer
GL(n, p ) = GL(y)
define a permutation
V
GL(n, p)
then
onto
[L, a]
[L, a](b) = L(b) + a , for all
denotes the identity of
p
n-dlmensional vector space over the field
V* = {a € V | a # 0} .
group of non-singular linear transformations of
L i GL(n, p )
55
V
of
b d V .
[J, a]
is the
then for each
7
by the
In particular, if
is simply translation
a . With this notation we now state and prove Theorem 1.
THEOREM
1. Let
T
\T\ = pn .
be a symmetric tournament,
may identify the points of
T
with those of
V(n, p )
Then we
so that
{[J, a] | a € V] 5 A(2") < {[L, a] \ I € GL(7), a i V] .
Further, identifying
T with V in this manner and setting
0(0) = {a (. V \ 0 > a] and 1(0) = {a € V | a > 0} tften the o r i i t s on V*
of A(T)Q , t/ie stabilizer of 0 in A(lT) , are 0(0) and 1(0) .
Moreover 0(0) = -1(0) .
Proof.
The first statement is an immediate consequence of our Lemma 1
and of Satz 3.5 [3, Chapter II]. The second statement is immediate, since
A(T)
is transitive on the arcs of
observe that
and
|I(0)| = (pn-l}/2 .
|0(0)| = (p -l)/2 , so
-a £ 0(0) . Since
a contradiction.
REMARK.
to
T . To prove the last statement
n
Thus
[J, -a] € A(T) we find
-a > 0
Suppose both
and
a
0 > -a ,
0(0) = -1(0) .
To fix notation let A(T) = G . Then the restrictions of G
0(0) or
1(0) yield representations
T
and
!„
respectively of G
as a transitive permutation group.
LEMMA
2.
The representations
representations of
Proof.
elements of
G
T
and
x
are faithful
and are similar.
Since both
0(0) and
1(0) contain half the non-zero
V , each contains a basis.
it follows that
Since
V
representations of
G . The last part of the lemma is proved by
0(0) onto
and
consists of linear
transformations of
considering the map from
T
G
x
1(0) given by
are faithful
a •* -a , and the map
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56
from
J.L. Berggren
T (G)
to
T (G)
that the elements of
given by
G
act linearly on
We next show that
irreducibly on
V .
T.(g) •*• T-(g) .
G
(Again one must recall
V .)
has a cyclic normal subgroup which acts
We first remark that any finite group
A
has a
maximal normal nilpotent subgroup, called its Fitting subgroup and written
Fit (A) .
LEMMA
3.
If
semi-regutarly on
Proof.
As
G = HT)
then Fit(G)
V* .
Fit(G)
is nilpotent it is a direct product of its Sylow
subgroups, so it suffices to show that if
Fit(G)
then
Q
G
0(0)
orbit of
How
0(0)
Q
on
0(0)
then
|Q| is odd so, as
Proof.
4.
on
V*
H
Q
on
is
Then
H
1(0) .
Aut(tf) , and, as
H
G
Q
H
U
W
is an
Thus
V* .
V* .
acts irreducibly on
(\n\, p) = 1 .
and
U
is
Q
V* , it follows from [4,
\H\ .
Now suppose
are proper
V .
V* , that
In particular,
H
(|#|, p)
acts
=
1 ) to
^-invariant subspaces
n
is odd, this means
acts semiregularly on
\H\ < \u\ = p™ .
But
V*
and
U
of
H . Thus
G/H
contains an
H = Fit(G) , and since
solvable it follows from Satz U.2(b) [3, Chapter III] that
the centralizer in
is a
and assume without loss of generality that
H
V* ,
0(0) , so
acts semiregularly on
m 5 [n/2] , and since
Since
Q
Q < G .
l/2-transitive on
We may apply Maschke's Theorem (as
Thus
on
Q
all have cardinality
\H\ (pn-l) ,
\ll\ = p
\l/\ S \w\ .
V* ,
is l/2-transitive on
V = U © W , where
Let
Since
is cyclic and acts semiregularly on
H = Fit(C) .
H
V .
m £ (n-l)/2 .
orbit of
is an orbit of
We have just shown that
since this means
V .
-U
Q
Q
Let
is, the orbits of
reducibly on
V* .
Fit(G) < G , we conclude
Ely the proof of Lemma 2, if
as a permutation group on
THEOREM
of
and
[4, Proposition h.kl, that is, the orbits of
Proposition 9-l6] that
conclude
Fit(G)
all have the same length.
viewing. Q
is a Sylow subgroup of
acts faithfully as a transitive group on
1/2-transitive on
on
Q
is cyclic and acts semiregularly on
characteristic subgroup of
By Lemma 2,
is cyclic and acts
H 2: CAH)
G
is
,
is isomorphic to a subgroup of
is cyclic of odd order,
|Aut(ff)| < \H\ .
Thus
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Finite symmetric tournaments
\G/B\ < \B\ .
orbit of
G
|c| = \G/U\\B\ < pmpm
Hence
and
|0(O)| = (p"-l}/2 .
contradiction shows
H
57
< p""1 .
Hence
acts irreducibly on
But
0(0)
n
(p -l)/2 < p * "
1
is an
.
This
V , for we may suppose
Pn>3.
Now observe that
group on
V
(1)
A(T)
is represented as a primitive permutation
with the following properties:
N = {[I, a] | a € V]
is an abelian minimal normal subgroup of
A(2") acting regularly on
(2)
V ;
A ( T ) 0 has the abelian normal subgroup
acts irreducibly on
B = Fit(A(r)Q)
V .
It is easy to show that, under these hypotheses,
subgroup of
BN .
which
N
is a minimal normal
The following theorem is an immediate consequence of
these remarks and Satz 3.12 [4, Chapter II].
THEOREM
5.
Let
T
the automorphism group of
1 .
n
GF(p )
n
p
,
n
GF(p ) , so that
T (pn) , the group of permutations of
a •*• xo(a) + a
where
3
c
Moreover,
translations of
GF(p n )
runs over the elements of
[the non-zero elements of
automorphisms.
and
Then we may identify the points of
the points of the Galois field of order
subgroup of
pn
be a symmetric tournament of order
n
GF(p ) ), and
T
with
A(T)
is a
of the form
GF(p ) ,
o
A(T)
x
runs over
runs over the field
A(I') contains the subgroup consisting of all
G?(p ) , permutations of the form
a •* a + c ,
n
a € GF(p ) .
REMARK.
A(7)
The last sentence is easy to show once one observes that
contains an abelian minimal normal subgroup acting regularly on
The permutation of
we shall denote by
notice, since
THEOREM
which maps an arbitrary
[x, a, c] . Further, let
p" = 3 C O
6.
GF(p )
Either
, S u (-5) = GF(pn}*
0(0) = S
or
a
to
xo{a) + c
S = {x2 | x € GF(p") }
while
T .
and
5 n (-S) = (3 .
0(0) = -5 .
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58
J.L. Berggren
Proof.
As
10(0) | = (p -l}/2 , if the theorem is false then there are
x, y $ GF(p n }
such that
[w, a, 0] = ir € A (2 I ) Q
there is
to = - L o ^ x " 1 ) !
< TT>
0 > x
2
= < IT > .
TT°(l) = 1
.
Let
S
•
Since
be t h e o r b i t of
2
and
TT(X 2 } = -y2
such that
u = Wx"1)
let
0 > -y2 . By arc transitivity
and
2
|A(T) Q |
is odd,
1
under the a c t i o n of
2
2
TT (l) = wa(u>) = - u a ( - u ) = (wa(u))
= y2 .
, that is,
.
?
< TTT > .
Now
Suppose
Then
= L o ( uO ( tf) )j
Thus the orbit of 1 under the action of < IT >> is a subset of
2
< TT > = < TT ) the same must be true of the orbit of 1 under < TT>
2
particular,
ir(l) = -u (l) = -u
(-1) = (su'1)2
REMARK.
2"
T .
, so
If, in
T ,
Clearly
T
1 ,
and
T'
0(0) = {x
GF (pn)
5 , so
\ x i GF(p ) } .
some
a > b
both in
S
.
Hence
(pn-l)/2
in
T'
iff
-a > -b
A(T)
is a subgroup
a > b
iff
K
of
0(0) = {x
K
T (j?n)
or both in
in
and
-S
b - a
iff
and
.
T
By
\ x (. GF (p
) } .
contains all
0 > b - a , that is,
b - a = x
To prove the last statement is now easy, for
T(p )
is odd.
We may now prove Theorem A.
then, since by Theorem 5,
r(p},
x € GF(p }
a subgroup of
.
^y Theorem 5 we may identify the points of
so that
a, b € GF(p )
translations in
ir(l) = z
then we may define a new tournament
by the rule
Theorem 6 and the remark after it we may assume
Now if
In
are isomorphic tournaments so we may suppose
Proof of Theorem A.
with those of
is in
0(0) = -5
GF(p n )
.
Since
2
It | (p"-l) , a contradiction since
on the points of
that, in
2
5 .
A(T)
for
is
[x, a, a](b) - [x, a, c](.a) are
x t S .
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Finite symmetric tournaments
59
References
[J]
Walter Feit and John G. Thompson, "Solvability of groups of odd
order", Pacific J. Math. 13 (1963), 775-1029.
[2]
Myron Goldberg, "The group of the quadratic residue tournament",
Canad. Math. Bull. 13 (1970), 51-51*.
[3]
B. Huppert, Endliche Gruppen I (Die Grundlehren der mathematischen
Wissenschaften, Band 134.
Springer-Verlag, Berlin, Heidelberg,
New York, 1967).
[4]
Donald Passman, Permutation groups (W.A. Benjamin, New York,
Amsterdam, 1968).
Simon Fraser University,
Burnaby,
British Columbia,
Canada.
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