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1. Mathematical Theory and Modeling
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.3, No.11, 2013
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Suborbits and Suborbital Graphs of the Symmetric Group Sn
acting on Ordered r-element Subsets
Jane Rimberia1*, Ireri Kamuti1, Bernard Kivunge1, Francis Kinyua2
Mathematics Department, Kenyatta University, P. O. Box 43844-00100, Nairobi, Kenya
Pure and Applied Mathematics Department, Jomo Kenyatta University of Agriculture and Technology,
Kenya P.O. BOX 62000, Nairobi, Kenya
*Email of the corresponding author: janekagwiria@yahoo.com
1.
2.
Abstract
[r]
The ranks and subdegrees of the symmetric group Sn acting on X , the set of all ordered r-element subsets
from X have been studied (See Rimberia [4]). In this paper, we examine some properties of suborbits and
[r]
suborbital graphs of Sn acting on X .
Keyword: Suborbital graphs, Suborbits, Symmetric Group, Group action
1.
Introduction
Suppose G is a group acting transitively on the set X and let Gx be the stabilizer in G of a point x ∈ X . The
∆ = { x} , ∆1 , ∆ 2 , K , ∆ k −1
orbits 0
of Gx on X are known as suborbits of G.
Now, let
( G, X ) be a transitive permutation group. Then G acts on
g ( x, y ) = ( g ( x ) , g ( y ) ) , g ∈ G , x, y ∈ X
.
X×X
The orbits of this action are called suborbitals of G. The orbit containing
Ο⊆ X ×X
{
x∈ X, ∆ = y∈ X
by
( x, y )
( x, y ) ∈ Ο}
is denoted by
Ο ( x, y )
. Now, if
is a G-orbit, then for any
is a Gx -orbit on X. Conversely, if
Ο = {( gx, gy ) g ∈ G , y ∈ ∆}
∆⊆ X
is a Gx -orbit, then
is G-orbit on X × X (Neumann [3]).
G ( x, y )
we can form a suborbital graph
: its vertices are elements of X, and there is a directed
x, y ) ∈ Ο ( x, y )
Ο ( y, x )
(
edge from x to y if and only if
. Clearly
is also a suborbital, and it is either equal
Ο ( x, y )
G ( x, y ) = G ( y, x )
to or disjoint from
. In the former case,
and the graph consists of pairs of oppositely
directed edges. It is convenient to replace each such pair by a single undirected edge, so that we have an
G ( x, y )
G ( y, x )
is just
with arrows reversed,
undirected graph which we call self-paired. In the latter case,
G ( x, y )
G ( y, x )
and we call
and
paired suborbital graphs.
From
Ο ( x, y )
The above ideas were first introduced by Sims [5], and are also described in a paper by Neumann [3] and in
books by Tsuzuku [6] and Biggs and White [1], the emphasis being on applications to finite groups.
Ο ( x, x ) = {( x, x ) x ∈ X }
x=y
If
, then
is the diagonal of X × X . The corresponding suborbital graph
G ( x, x )
, called the trivial suborbital graph, is self-paired and consists of a loop based at each vertex x ∈ X .
2. Notations and Preliminary Results
Notation 2.1
[r ]
Throughout this paper, G is the symmetric group S n while X and
X = {1, 2, K , n}
and a suborbital graph respectively.
subsets from
115
G denotes the set of all ordered r-element

2. Mathematical Theory and Modeling
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.3, No.11, 2013
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Definition 2.1
If σ ∈ S n has α1 cycles of length 1, α 2 cycles of length 2, …, α n cycles of length n, then the cycle type of σ is
(α , α , K , α n ) .
the n-tuple 1 2
Definition 2.2
∆ ∗ = { gx g ∈ G , x ∈ g ∆}
∗
Let G act transitively on a set X and let ∆ be an orbit of Gx on X. Define
, then ∆ is
∆∗∗ = ∆
also an orbit of Gx and is called the Gx -orbit (or G-suborbit) paired with ∆ (Wielandt [7]). Clearly
∆ = ∆∗
∆ = ∆∗
and
. If
, then ∆ is said to be self-paired.
Definition 2.3
Let G act on a set X, then the character π of permutation representation of G on X is defined by
π ( g ) = Fix ( g ) , for all g ∈ G
.
Theorem 2.1 (Cameron [2])
Let G act transitively on a set X, and let g ∈ G . Suppose π is the character of the permutation representation of
G on X, then the number of self-paired suborbits of G is given by
1
nπ =
∑ π g2
G g ∈G
.
Theorem 2.2 (Sims [5])
( )
∆ = { x} , ∆1 , ∆2 , K , ∆k −1
G
Let G be transitive on X and let x be the stabilizer of the point x ∈ X . Suppose 0
are
Gx on X and let Οi ⊆ X × X , i = 0, 1, K , k − 1 be the suborbital corresponding to
the orbits of
∆ i , i = 0, 1, K , k − 1 . Then G is primitive if and only if each suborbital graph Gi , i = 1, 2, K , k − 1 is connected.
[r ]
3.
Properties of Suborbits of G acting on X
Theorem 3.1
G
[r ]
[r ]
∆ = [ a1 , a2 , K , ar ]
Let G act on X . Suppose
is an orbit of [1, 2, ... , r ] on X
of length 1, where
 1 2 ... r 
σ =

ai ∈ {1, 2, K , r} i = 1, 2, K , r
 a1 a2 ... ar  is such
,
. Then ∆ is self-paired if and only if the permutation
σ 2 = 1.
that
Proof
Let ∆ be self-paired. Then there exists g ∈ G such that
g [ a1 , a2 , K , ar ] = [1, 2, K , r ]
, that is
 g ( a1 ) , g ( a2 ) , K , g ( ar )  = [1, 2, K , r ]


⇒
g ( a1 ) = 1, g ( a2 ) = 2,..., g ( ar ) = r
.
.
∆ is self-paired, then by Definition 2.2
Since
g (1) = a1 , g ( 2 ) = a2 ,..., g ( r ) = ar
.
⇒ g exchanges ai and i if ai ≠ i or fixes i . Thus the permutation
 1 2 ... r 
σ =

 a1 a2 ... ar 
σ 2 = 1. Conversely, let σ 2 = 1 , then σ = σ −1 . Now, g ∈ G such that
is such that
 1 2 ... r ... n 
g =

 a1 a2 ... ar ... an  takes [ a1 , a2 , K , ar ] to [1, 2, K , r ] and [1, 2, K , r ] to [ a1 , a2 , K , ar ] . Therefore ∆ is

self-paired.
Theorem 3.2
[r ]
∆ = [ a1 , a2 , K , ar ]
∆ = [b1 , b2 , K , br ]
a , b ∈ {1, 2, K , r}
Let G act on X and suppose i
and j
, where i i
,
116

3. Mathematical Theory and Modeling
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.3, No.11, 2013
i = 1, 2, K , r
are orbits of
G[1, 2, ... , r ]
www.iiste.org
[r ]
∆
on X of length 1. Then ∆ i and j are paired if and only if the
permutations
 1 2 ... r 
 1 2 ... r 
σi = 
σj =


 b1 b2 ... br  and
 a1 a2 ... ar  are inverses of each other.
Proof
∆
g , g ∈G
Suppose ∆ i and j are paired. Then there exist i j
such that
gi [ a1 , a2 , K , ar ] = [1, 2, K , r ]
g [b , b , K , br ] = [1, 2, K , r ]
and j 1 2
.
That is,
 g ( b ) , g j ( b2 ) , ..., g j ( br )  = [1, 2, K , r ]
 g i ( a1 ) , g i ( a2 ) , K , gi ( ar )  = [1, 2, K , r ]



and  j 1
.
⇒ gi ( a1 ) = 1, gi ( a2 ) = 2, K , gi ( ar ) = r and g j ( b1 ) = 1, g j ( b2 ) = 2, K , g j ( br ) = r .
∆
Since ∆ i and j are paired, then by Definition 2.2
gi (1) = b1 , gi ( 2 ) = b2 , K , gi ( r ) = br
g (1) = a1 , g j ( 2 ) = a2 , K , g j ( r ) = ar
and j
.
⇒ ( g i g j ) (1) = 1, ( gi g j ) ( 2 ) = 2, K , ( g i g j ) ( r ) = r .
Similarly,
( g j gi ) (1) = 1, ( g j gi ) ( 2 ) = 2, K , ( g j gi ) ( r ) = r .
 1 2 ... r 
 1 2 ... r 
σi = 
σj =


 b1 b2 ... br  and
 a1 a2 ... ar  are inverses of each other. Conversely,
Hence the permutations
 1 2 ... r 
 1 2 ... r 
σi = 
σj =


 b1 b2 ... br  and
 a1 a2 ... ar  are inverses of each other. Now, if gi , g j ∈ G where
suppose
 1 2 ... r ... n 
 1 2 ... r ... n 
gi = 
gj = 


 b1 b2 ... br ... bn  and
 a1 a2 ... ar ... an  , then g i takes [ a1 , a2 , K , ar ] to [1, 2, K , r ] and
[1, 2, K , r ] to [b1 , b2 , K , br ] . Similarly, g j takes [b1 , b2 , K , br ] to [1, 2, K , r ] and [1, 2, K , r ] to
[ a1 , a2 , K , ar ] . Hence ∆ i and ∆ j are paired.

Lemma 3.1
(α , α , K , α n ) . If α1 ≥ r , then the number of elements in X [r ] fixed by g is
Let the cycle type of g ∈ G be 1 2
given by
α 
Fix ( g ) = r ! 1 
 r .
3.1)
Proof
[ a , a , K , ar ] ∈ X [r ] and g ∈ G . Then g fixes [ a1 , a2 , K , ar ] ∈ X [r ] if and only if a1 , a2 , .., ar are mapped
Let 1 2
onto themselves by g. That is
g [ a1 , a2 , K , ar ] =  g ( a1 ) , g ( a2 ) , K , g ( ar )  = [ a1 , a2 , K , ar ] implying that


g ( a1 ) = a1 , g ( a2 ) = a2 , K , g ( ar ) = ar
. Therefore each of the elements a1 , a2 , K , ar comes from a 1-cycle in
 α1 
 
r
g. Hence the number of unordered r-element subsets fixed by g ∈ S n is   . But an unordered r-element
{a , a , K , ar } can be rearranged to give r ! distinct ordered r-element subsets. Hence
subset say, 1 2
α 
Fix ( g ) = r ! 1 
 r .

Theorem 3.3
[r ]
(α , α , K , α n ) , then the number of self-paired suborbits
Let G act on X and suppose g ∈ G has cycle type 1 2
117

4. Mathematical Theory and Modeling
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.3, No.11, 2013
[r ]
of G on X is given by
r !  α + 2α 2 
nπ = ∑  1

r
n! g 

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.
(3.2)
Proof
(α , α , K , α n ) ,, then g 2 has (α1 + 2α2 ) cycles of length 1. Hence by Lemma
Let the cycle type of g ∈ G be 1 2
[r ]
g2
3.1, the number of elements in X fixed by
is given by
α1 + 2α 2 

Fix g 2 = r !

r

.
( )
[r ]
By using this together with Theorem 2.1 we see that the number of self-paired suborbits of G on X is equal to
1
 α + 2α 
1
nπ =
∑G π g 2 = n ! ∑ r ! 1 r 2 
G g∈
g


( )
=
4.
r !  α1 + 2α 2 
∑ r 
n! g 


.
[r ]
Suborbital Graphs of G acting on X
[r ]
Construction of Suborbital Graphs of G acting on X
G
[r ]
[r ]
[ a , a , K , ar ] ∈ ∆ , where
Let G act on X and let ∆ be an orbit of [1, 2, ... , r ] on X . Suppose 1 2
ai ∈ {1, 2, K , n} i = 1, 2, K , r
,
. Then the suborbital Ο corresponding to ∆ is given by
4.1
Ο=
{( g [1, 2, K , r ] , g [ a , a
1
2
, K , a r ])
g ∈ G , [ a1 , a 2 , K , a r ] ∈ ∆
}.
[r ]
We form the suborbital graph G corresponding to suborbital Ο by taking X as the vertex set and by
[b , b , K , br ] to [c1 , c2 , K , cr ] if and only if ([b1 , b2 , K , br ] , [c1 , c2 , K , cr ]) ∈ Ο .
including an edge from 1 2
[1, 2, K , r ] in positions i, j, k , K are respectively identical to the coordinates of
Now, if the coordinates of
[ a1 , a2 , K , ar ] in positions x, y , z , K ., then ([b1 , b2 , K , br ] , [c1 , c2 , K , cr ]) ∈ Ο if and only if the
[b , b , K , br ] in positions i, j, k , K are respectively identical to the coordinates of
coordinates of 1 2
[c1 , c2 , K , cr ] in positions x, y , z , K . Consequently we have an edge from [b1 , b2 , K , br ] to [c1 , c2 , K , cr ] in
G .
[r ]
4.2
Properties of Suborbital Graphs of G acting on X
Lemma 4.2.1 (Rimberia [4])
[r ]
The action of G on X is imprimitive if n > r + 1 .
Theorem 4.2.1
[r ]
If n > r + 1 , then all the suborbital graphs corresponding to the action of G on X are disconnected.
Proof
[r ]
By Lemma 4.2.1, G acts imprimitively on X if n > r + 1 hence by Theorem 2.2 all the corresponding

suborbital graphs are disconnected provided n > r + 1.
Next, we consider the other two cases:
Case 1
G
[r ]
n=r
If
, then each orbit of [1, 2,..., r ] on X is of length 1. Thus the suborbital graphs corresponding to the selfpaired suborbits are regular of degree 1 and so must be disconnected. On other hand, the suborbital graphs
corresponding to the paired suborbits have vertices each of which has indegree 1 and outdegree 1. Furthermore,
any two consecutive vertices S and T in these graphs need not be adjacent since for there to be a directed edge,
say from S to T, the coordinates of S and T must satisfy the rule defining the corresponding suborbital. Therefore
118

5. Mathematical Theory and Modeling
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.3, No.11, 2013
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such a graph cannot have a directed cycle containing every vertex, and so must be disconnected.
Case 2
G
( r − 1) elements from A = {1, 2, K , r} . Now, the
If n = r + 1 , then [1, 2,..., r ] has orbits with exactly r and
n − r = ( r + 1) − r = 1
former orbits have length 1 while the latter have length
(Rimberia [4]). Thus the
corresponding suborbital graphs have vertices each of which has degree 1 or indegree 1 and outdegree 1.
Similarly, these graphs must be disconnected.
Theorem 4.2.2
[r ]
Let G act on X . Then the number of connected components in the suborbital graph Gi corresponding to a selfG 1,2,...,r ]
[r ]
A = {1, 2,..., r}
paired orbit of [
on X with exactly r elements from
is equal to
n!
n ( Gi ) =
2 ( n − r )!
.
4.1)
Proof
G
[r ]
Let Gi be the suborbital graph corresponding to a self-paired orbit of [1, 2, K , r ] on X with exactly r elements
A = {1, 2, K , r}
from
. Since each vertex of Gi has degree 1, then the connected components in Gi are trees with
two vertices and one edge. Hence the number of connected components in Gi is equal to
n
r
n!
  r!
X[ ]
Number of vertices in Gi
=
r
n ( Gi ) =
=
= 
2 (n − r )!
2
2
2

.
Theorem 4.2.3
[r ]
G
Let G act on X . Then the number of connected components in the suborbital graph j corresponding to a
G 1,2,...,r ]
[r ]
A = {1, 2,..., r}
paired orbit of [
on X with exactly r elements from
is equal to
n!
n (Gj ) =
3 (n − r )!
.
4.2)
Proof
G
[r ]
G
Let j be the suborbital graph corresponding to a paired orbit of [1 ,2, K , r ] on X with exactly r elements from
A = {1, 2, K , r}
G
. Then each vertex of j has indegree 1 and outdegree 1. Moreover, construction shows that the
G
G
connected components of j are directed triangles. Hence the number of connected components in j is equal to
n
r
n!
X[ ]
  r!
Number of vertices in G j
=
r
n (Gj ) =
=
= 
3 ( n − r )!
3
3
3

.
Corollary 4.2.1
G
[r ]
[r ]
Let G act on X and let Gi be the suborbital graph corresponding to a self-paired orbit of [1, 2, K , r ] on X
A = {1, 2, K , r}
with exactly r elements from
. Then Gi has girth equal to zero.
Proof
By Theorem 4.2.2, the connected components in Gi are trees with two vertices and one edge. Hence Gi cannot

have a cycle which implies that its girth is equal to zero.
Corollary 4.2.2
G
[r ]
[r ]
G
Let G act on X and let j be the suborbital graph corresponding to a paired orbit of [1, 2, K , r ] on X with
A = {1, 2, K , r}
G
exactly r elements from
. Then j has girth 3.
Proof
G
By Theorem 4.2.3, the connected components in j are directed triangles, that is, directed cycles of length 3.
G

Hence the girth of j is equal to 3.
119

6. Mathematical Theory and Modeling
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.3, No.11, 2013
www.iiste.org
Theorem 4.2.4
G
[r ]
[r ]
Let G act on X and let G be the suborbital graph corresponding to the orbit of [1, 2, K , r ] on X with no
A = {1, 2,..., r}
element from
. Then G has girth 3 provided n ≥ 3r .
Proof
G[1, 2, K , r ]
[r ]
A = {1, 2, K , r}
Let ∆ be the orbit of
on X
with no element from
and suppose
d1 , d 2 , K , d r ] ∈ ∆
[
. Then the suborbital Ο corresponding to ∆ is given by
Ο=
{( g [1, 2, K , r ] , g [d , d
1
2
}.
, K , d r ] ) g ∈ G , [ d1 , d 2 , K , d r ] ∈ ∆
Therefore the corresponding suborbital graph
G has X [
r]
as the vertex set and has an edge from
[ e1 , e2 , K , er ] to [ f , f , K , f ] if and only if {e , e , K , e } ∩ { f , f
1
4.1 below exists in
2
1
r
2
r
1
2
, K , fr } = φ
G if and only if the sets {d1 , d2 , K , dr } , {e1 , e2 , K , er }
and
disjoint. But clearly this is possible if n ≥ 3r .
. Hence the cycle in Figure
f2 , K , fr }
are mutually
{ f1 ,
[ f1 , f2 , K , fr ]
[e1 , e2 , K , er ]
[ d1 , d2 , K , dr ]
Figure 4.1: A cycle in G

References
[1]
Biggs, N.M. and White A. T. (1979). Permutation Groups and Combinatorial Structures. London
Maths. Soc. Lecture Notes 33, Cambridge University Press, Cambridge.
[2]
Cameron, P. J. (1974). Suborbits in transitive permutation groups. Proceedings of the NATO Advanced
Study Institute on Combinatorics, Breukelen, Netherlands.
[3]
Neumann, P. M. (1977). Finite permutation groups, edge-coloured graphs and matrices. In Curran M. P.
(Ed.), Topics in Group Theory and Computation, Academic Press, London.
[4]
[5]
[6]
[7]
S
Rimberia, J. K. (2011). Rank and subdegrees of the symmetric group n acting on ordered r-element
subsets. Proceedings of the First Kenyatta University International Mathematics Conference held
between 8th-10th June, 2011 in Nairobi, Kenya.
Sims, C. C. (1967). Graphs and finite permutation groups. Math. Zeitschrift 95: 76 – 86.
Tsuzuku, T. (1982). Finite Groups.and Finite Geometries, Cambridge University Press, Cambridge.
Wielandt, H. (1964). Finite permutation groups. Academic Press, New York
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