1 Introduction
This article deals with computing π-th powers in the wreath product
G
β
S
n
, where πΊ is a finite group and
r
β₯
2
is an integer.
In some form or the other, these groups occur as natural subgroups of the symmetric group such as the centralizer of elements, normalizer of certain subgroups, Sylow subgroups and, as such, play an important role in the theory of permutation groups.
The group
C
m
β
S
n
is called the generalized symmetric group, where
C
m
is the cyclic group of order π.
Alternatively,
C
m
β
S
n
also appears in the classification of complex reflection groups (see [10, Theorem 8.29]) where it is denoted by
G
β’
(
m
,
1
,
n
)
.
In particular, the group
C
2
β
S
n
is called the hyper-octahedral group.
These groups are also examples of Weyl groups.
The Weyl groups of type
A
n
,
B
n
,
C
n
are wreath products, namely
S
n
+
1
,
C
2
β
S
n
,
C
2
β
S
n
respectively.
This investigation of powers in the group
G
β
S
n
is motivated by the study of powers in
S
n
, in which a substantial amount of information is available.
When πΊ is trivial,
G
β
S
n
is
S
n
itself, and therefore our results in this article generalize some of the known results obtained for powers in
S
n
, for example in [4, 6] where the authors have studied the proportion of π-th powers in
S
n
.
Let
r
β₯
2
be any integer.
The power map
Ο
r
:
G
β
S
n
β
G
β
S
n
is defined by
Ο
r
β’
(
g
)
=
g
r
for every
g
β
G
β
S
n
.
The elements in the image
Ο
r
β’
(
G
β
S
n
)
=
{
g
r
β£
g
β
G
β
S
n
}
are called π-th powers in
G
β
S
n
.
Let
P
r
β’
(
G
β
S
n
)
:=
|
Ο
r
β’
(
G
β
S
n
)
|
|
G
|
n
β’
n
!
denote the probability that a randomly chosen element of
G
β
S
n
is an π-th power.
In this article, we give a characterization of π-th powers in
G
β
S
n
, assuming π is a prime, and thus obtain a generating function for
P
r
β’
(
G
β
S
n
)
.
We use this generating function to prove a partial recursive relation as follows: for a prime
r
β₯
2
with
(
r
,
|
G
|
)
=
1
,
P
r
β’
(
G
β
S
n
)
=
P
r
β’
(
G
β
S
n
+
1
)
if
r
β’
-
1
β’
(
mod
β’
r
)
.
This is stated as Theorem 5.6.
When πΊ is trivial, we get back this well-known relation for
S
n
(see [6, Theorem 3.4]).
The set of π-th powers
Ο
r
β’
(
G
β
S
n
)
is closed under conjugation and therefore is a union of conjugacy classes of
G
β
S
n
.
A conjugacy class of
G
β
S
n
contained in
Ο
r
β’
(
G
β
S
n
)
is called an π-th power conjugacy class.
A formula is deduced for the number of π-th power conjugacy classes of
G
β
S
n
in terms of the function
p
β’
(
n
)
, which denotes the number of partitions of a natural number π.
We state and prove this in Theorem 5.9.
As stated earlier, the results proved here bring the results obtained for the symmetric group to a more general context.
It is worth mentioning that the power map along these lines has been studied for the group
GL
β’
(
n
,
q
)
, which is the group of all
n
Γ
n
invertible matrices over the finite field with π elements, by the first author and A. Singh in [9].
In another direction, the authors in [8] showed that asymptotically (more precisely, for large field size), the proportion of powers in finite reductive groups can be essentially computed inside their Weyl groups.
The paper is organized as follows.
Section 2 collects some of the important results on powers in the symmetric group.
This section serves as a short survey on the topic and also allows us to point out the specific generalizations that we obtain for powers in
G
β
S
n
.
In Section 3, we discuss the notion of cycle index in
G
β
S
n
, which is the generalization of PΓ³lyaβs cycle index for
S
n
.
The notion of cycle index will play a major role in this paper.
Section 4 is the main technical section where we compute the powers in
G
β
S
n
and characterize them (see Proposition 4.3).
In Section 5, we write a generating function for the proportion of powers in
G
β
S
n
(see Proposition 5.1) and prove one of the main results, which is Theorem 5.6.
A formula for the number of π-th power conjugacy classes is given in Theorem 5.9.
We end this paper with a short discussion on the asymptotics of the powers in
G
β
S
n
and mention some interesting questions in this direction.
2 Powers in symmetric group
Here, we collect some well-known results on powers in symmetric groups.
Let
r
β₯
2
be an integer.
Consider the power map
Ο
r
:
S
n
β
S
n
defined by
Ο
r
β’
(
Ο
)
=
Ο
r
for all
Ο
β
S
n
.
The image of this map,
Ο
r
β’
(
S
n
)
=
{
Ο
r
β£
Ο
β
S
n
}
, is the set of π-th powers in
S
n
.
Let
P
r
β’
(
S
n
)
:=
|
Ο
r
β’
(
S
n
)
|
n
!
be the probability that a randomly chosen element in
S
n
is an π-th power.
For a natural number π,
Ξ»
=
(
Ξ»
1
,
Ξ»
2
,
β¦
)
,
where
β’
Ξ»
i
β₯
0
β’
for all
β’
i
β’
and
β’
Ξ»
1
β₯
Ξ»
2
β₯
β―
,
is called a partition of π if
β
i
Ξ»
i
=
n
.
We denote a partition π of π by
Ξ»
β’
n
.
The positive integers
Ξ»
i
are called the parts of π.
In power notation, we write
Ξ»
=
1
m
1
β’
2
m
2
β’
β¦
β’
i
m
i
β’
β¦
β
with
β’
β
i
i
β’
m
i
=
n
,
where
m
i
denotes the number of times π occur as a part in π.
The conjugacy classes in the group
S
n
are given in terms of the cycle structure of permutations.
Any permutation
Ο
β
S
n
can be written as a product of disjoint cycles; thus we define
type
β‘
(
Ο
)
=
(
c
1
,
c
2
,
β¦
)
, where
c
i
denotes the number of π-cycles in π.
It satisfies the relation
β
i
i
β’
c
i
=
n
and thus actually defines a partition of π, that in power notation is
1
c
1
β’
2
c
2
β’
β¦
β’
n
.
It is well known that two permutations
Ο
,
Ο
β
S
n
are conjugate if and only if
type
β‘
(
Ο
)
=
type
β‘
(
Ο
)
.
Thus, conjugacy classes in
S
n
are in one-to-one correspondence with partitions of π.
The π-th powers in
S
n
can be computed easily if we assume that π is a prime.
The following lemma, which can be found in [6], will also be useful for our computation in the wreath product.
We include a proof since it will be required later.
Lemma 2.1
Lemma 2.1 ([6, Proposition 3.1])
Let
r
β₯
2
be a prime.
Let
Ο
β
S
n
be of type
(
c
1
,
c
2
,
β¦
)
.
Then
type
β‘
(
Ο
r
)
=
(
d
1
,
d
2
,
β¦
)
, where
d
i
=
{
c
i
+
r
β’
c
r
β’
i
when
β’
r
β€
i
,
r
β’
c
r
β’
i
otherwise
.
Proof
Suppose πΌ is a π-cycle and
r
β€
k
.
Then it is easy to observe that
Ξ±
r
is a π-cycle.
If
r
β£
k
, say
k
=
a
β’
r
, then once again it is easy to see that
Ξ±
r
decomposes as a product of π number of π-cycles.
Assume now that
Ο
β
S
n
has type
(
c
1
,
c
2
,
β¦
)
and
Ο
r
has type
(
d
1
,
d
2
,
β¦
)
.
Let
1
β€
i
β€
n
be such that
r
β£
i
.
Since
d
i
denotes the number of π-cycles in
Ο
r
,
c
r
β’
i
denotes the number of
r
β’
i
-cycles in π and the fact that each
r
β’
i
-cycle when raised to the power of π gives π separate π-cycles, it is clear that
d
i
=
r
β’
c
r
β’
i
.
Observe that, in the above case, the
r
β’
i
-cycles are the only ones which contribute to the number of π-cycles in
Ο
r
.
Now, assume that
r
β€
i
.
In this case, apart from the
r
β’
i
-cycles which once again contribute π separate π-cycles each when raised to the power π, the π-cycles themselves in π remain π-cycles in
Ο
r
.
There are
c
i
number of π-cycles in π; whence if
r
β€
i
,
d
i
=
c
i
+
r
β’
c
r
β’
i
.
This completes the proof.
β
Using the above lemma, one can easily characterize the permutations which are π-th powers.
The following proposition can be found once again in [6].
We state it without proof.
Proposition 2.2
Let
r
β₯
2
be a prime.
Let
Ο
β
S
n
with
type
β‘
(
Ο
)
given by the partition
Ξ»
=
1
m
1
β’
2
m
2
β’
β¦
β’
i
m
i
β’
β¦
β’
n
.
Then
Ο
β
Ο
r
β’
(
S
n
)
if and only if
r
β£
m
i
whenever
r
β£
i
.
In [6], the authors showed that the probability of π-th powers in
S
n
, that is,
P
r
β’
(
S
n
)
, satisfies the following property.
Theorem 2.3
Theorem 2.3 ([6, Theorem 3.4])
Let
r
β₯
2
be a prime.
Then
P
r
β’
(
S
n
)
=
P
r
β’
(
S
n
+
1
)
for all
n
β₯
0
and
n
+
1
β’
0
β’
(
mod
β’
r
)
.
Thus,
|
Ο
r
β’
(
S
n
+
1
)
|
=
(
n
+
1
)
β’
|
Ο
r
β’
(
S
n
)
|
when
n
β’
-
1
β’
(
mod
β’
r
)
.
Example 2.4
Observe that Theorem 2.3 does not hold if we take π to be composite.
Let
r
=
6
.
We have
P
6
β’
(
S
4
)
=
1
6
β
1
3
=
P
6
β’
(
S
5
)
.
The authors in [6] proved this result using bijective methods.
They further investigated
P
r
β’
(
S
n
)
as a sequence in π and proved that it is monotonically decreasing in π and
lim
n
β
β
β‘
P
r
β’
(
S
n
)
=
0
.
In other words, powers in the symmetric groups are really scarce for large π.
The particular case of
r
=
2
was investigated by J. Blum in [4], using generating functions.
The generating function for the powers in
S
n
can be derived using Polyaβs cycle index of
S
n
.
We briefly describe the notion of cycle index of
S
n
here (see [16] for more details), for we will see in the next section that this idea can also be generalized for
G
β
S
n
.
The cycle index of
S
n
is defined as
(2.1)
Z
n
=
Z
n
β’
(
t
1
,
t
2
,
β¦
,
t
n
;
S
n
)
=
1
n
!
β’
β
Ο
β
S
n
type
β‘
(
Ο
)
=
(
c
1
,
c
2
,
β¦
)
t
1
c
1
β’
t
2
c
2
β’
β¦
β’
t
i
c
i
β’
β¦
.
Observe that the product inside is a finite product since there exists
m
β
N
such that
c
i
=
0
for all
i
β₯
m
.
The coefficient of the monomial
t
1
a
1
β’
t
2
a
2
β’
β¦
,
where
β’
β
i
i
β’
a
i
=
n
,
is equal to
|
Cl
β‘
(
Ο
)
|
n
!
, where
Ο
β
S
n
is such that
type
β‘
(
Ο
)
=
(
a
1
,
a
2
,
β¦
)
, and
Cl
β‘
(
Ο
)
denotes the conjugacy class of π.
Example 2.5
The cycle index
Z
3
=
Z
3
β’
(
t
1
,
t
2
,
t
3
;
S
3
)
of
S
3
is given by
1
6
β’
(
t
1
3
+
3
β’
t
1
β’
t
2
+
2
β’
t
3
)
.
The cycle index generating function is given by
(2.2)
1
+
β
n
=
1
β
Z
n
β’
u
n
=
β
i
β₯
1
exp
β‘
(
t
i
β’
u
i
i
)
.
Using this, one can write the generating function for the proportion of square permutations (see [4, Equation 5]), which is
(2.3)
1
+
β
n
=
1
β
P
2
β’
(
S
n
)
β’
u
n
=
(
1
+
u
1
-
u
)
1
2
β’
β
k
=
1
β
cosh
β‘
(
u
2
β’
k
2
β’
k
)
.
Blum used analytic properties of this generating function to give asymptotic estimate of
P
2
β’
(
S
n
)
.
He proved that
P
2
β’
(
S
n
)
βΌ
K
β’
2
Ο
β’
n
-
1
2
,
where
β’
K
=
β
k
=
1
β
cosh
β‘
(
1
2
β’
k
)
.
In [15], the author investigated the powers in
S
n
for any
r
β₯
2
.
He generalized Blumβs estimate for
P
2
β’
(
S
n
)
as
P
r
β’
(
S
n
)
βΌ
n
β
β
Ο
r
n
1
-
Ο
β’
(
r
)
/
r
,
where π denotes Eulerβs phi function and
Ο
r
is an explicit constant.
In [1], the authors have determined the constant
Ο
r
more explicitly.
For more results and estimates on powers in
S
n
and related ideas, we urge the reader to look at [2, 5, 3, 11, 13, 17].
The squares in the alternating group
A
n
are determined in [14].
3 Cycle index of the wreath product
G
β
S
n
In this section, we will briefly discuss the notion of cycle index for the wreath product
G
β
S
n
for a finite group πΊ, as indicated in [12].
As mentioned before, this is a generalization of the cycle index for
S
n
(see equation (2.1)).
Using the cycle index, we will obtain generating functions for powers in
G
β
S
n
in Section 5.
We start by briefly describing the group
G
β
S
n
, in the form we will use.
We follow the exposition in [7].
Let πΊ be a finite group and π» a permutation group on
Ξ©
=
{
1
,
2
,
β¦
,
n
}
.
Define
G
β
H
=
{
(
f
,
Ο
)
β£
f
:
Ξ©
β
G
,
Ο
β
H
}
,
where
f
:
Ξ©
β
G
is any function.
Define multiplication on
G
β
H
as
(
f
,
Ο
)
.
(
f
β²
,
Ο
β²
)
=
(
f
β’
f
Ο
β²
,
Ο
β’
Ο
β²
)
,
where
f
Ο
:
Ξ©
β
G
is defined by
f
Ο
β’
(
i
)
=
f
β’
(
Ο
-
1
β’
(
i
)
)
for every
i
β
Ξ©
and for two functions
f
,
g
:
Ξ©
β
G
,
f
β’
g
:
Ξ©
β
G
is defined by
(
f
β’
g
)
β’
(
i
)
=
f
β’
(
i
)
β’
g
β’
(
i
)
.
Therefore,
f
β’
f
Ο
β²
:
Ξ©
β
G
is defined by
f
β’
f
Ο
β²
β’
(
i
)
=
f
β’
(
i
)
β’
f
β²
β’
(
Ο
-
1
β’
i
)
for every
i
β
Ξ©
.
The set
G
β
H
with the above binary operation forms a group with
(
id
,
1
H
)
as the identity, where
id
:
Ξ©
β
G
is defined by
id
β‘
(
i
)
=
1
G
for every
i
β
Ξ©
.
For
(
f
,
Ο
)
β
G
β
H
,
(
f
Ο
-
1
-
1
,
Ο
-
1
)
is its inverse, where
f
-
1
:
Ξ©
β
G
is defined by
f
-
1
β’
(
i
)
=
f
β’
(
i
)
-
1
for every
i
β
Ξ©
.
We have
f
Ο
-
1
-
1
=
(
f
Ο
-
1
)
-
1
.
This completes the description of the group
G
β
H
which is called πΊ wreath π».
We have
|
G
β
H
|
=
|
G
|
n
β’
|
H
|
.
The above description also shows that
G
β
H
is the semidirect product of
G
n
=
G
Γ
G
Γ
β¦
Γ
G
β
n
β’
times
with π», under the automorphism
Ο
:
H
β
Aut
β’
(
G
n
)
, where
Ο
h
acts on
G
n
by changing indices according to β.
For the rest of the paper, we take
H
=
S
n
.
To define the cycle index of
G
β
S
n
, we need the description of conjugacy classes in
G
β
S
n
.
We have already seen that conjugacy classes in
S
n
are characterized by the type of a permutation which is just a partition of π.
One can generalize this notion of the type of a permutation to understand the conjugacy classes in
G
β
S
n
(see [7]).
We describe it briefly.
Let
(
f
,
Ο
)
β
G
β
S
n
.
Let
(
j
,
Ο
β’
(
j
)
,
β¦
,
Ο
t
β’
(
j
)
)
be a cycle appearing in π (when written as a product of disjoint cycles).
Define the cycle product corresponding to that cycle in π as the unique element in πΊ given by
f
β’
(
j
)
β’
f
Ο
β’
(
j
)
β’
β¦
β’
f
Ο
t
β’
(
j
)
=
f
β’
(
j
)
β’
f
β’
(
Ο
-
1
β’
(
j
)
)
β’
β¦
β’
f
β’
(
Ο
-
t
β’
(
j
)
)
.
Let
C
1
,
C
2
,
β¦
,
C
s
denote a labeling of the conjugacy classes of πΊ.
We define the type of an element
(
f
,
Ο
)
β
G
β
S
n
as the
s
Γ
n
matrix
(
a
i
β’
k
)
, where
a
i
β’
k
is the number of π-cycles in π whose cycle product belongs to
C
i
.
If we denote
type
β‘
(
Ο
)
=
(
c
1
,
c
2
,
β¦
)
, then it is clearly seen that
a
i
β’
k
β
Z
β₯
0
for every
1
β€
i
β€
s
,
1
β€
k
β€
n
and
β
i
a
i
β’
k
=
c
k
for every
k
=
1
,
2
,
β¦
,
n
.
Moreover,
β
i
,
k
k
β’
a
i
β’
k
=
n
.
The following result that we state without proof determines the conjugacy classes of
G
β
S
n
and associates combinatorial data to each such conjugacy class.
Proposition 3.1
Proposition 3.1 ([7, 3.7])
Let
(
f
,
Ο
)
,
(
g
,
Ο
β²
)
β
G
β
S
n
.
Then
(
f
,
Ο
)
is conjugate to
(
g
,
Ο
β²
)
if and only if
type
β‘
(
f
,
Ο
)
=
type
β‘
(
g
,
Ο
β²
)
.
Therefore, the conjugacy classes in
G
β
S
n
are in one-to-one correspondence with
s
Γ
n
matrices
(
a
i
β’
k
)
with
a
i
β’
k
β
Z
β₯
0
for every
i
,
k
and
β
i
,
k
k
β’
a
i
β’
k
=
n
.
The type of
(
f
,
Ο
)
is well-defined in the sense that the cycle product does not depend on the representation of a cycle
(
j
,
Ο
β’
(
j
)
,
β¦
,
Ο
t
β’
(
j
)
)
in π.
In other words, if
(
j
,
Ο
β’
(
j
)
,
β¦
,
Ο
t
β’
(
j
)
)
=
(
l
,
Ο
β’
(
l
)
,
β¦
,
Ο
t
β’
(
l
)
)
, then their corresponding cycle products
f
β’
(
j
)
β’
f
Ο
β’
(
j
)
β’
β¦
β’
f
Ο
t
β’
(
j
)
and
f
β’
(
l
)
β’
f
Ο
β’
(
l
)
β’
β¦
β’
f
Ο
t
β’
(
l
)
are conjugate in πΊ.
Later, we will see that this in fact gives us a degree of freedom to choose a suitable cycle product in various computations.
Given
g
=
(
f
,
Ο
)
β
G
β
S
n
, let
Z
β’
(
g
)
denote the centralizer of π in
G
β
S
n
.
The structure of the centralizer is well known.
For our purpose, we will need the order of the centralizer.
Proposition 3.2
Proposition 3.2 ([7, 3.9])
Let
g
=
(
f
,
Ο
)
β
G
β
S
n
and
type
β‘
(
g
)
=
(
a
i
β’
k
)
which is an
s
Γ
n
matrix with
a
i
β’
k
β
Z
β₯
0
and
β
i
,
k
k
β’
a
i
β’
k
=
n
.
Then
|
Z
β’
(
g
)
|
=
β
i
,
k
a
i
β’
k
!
β’
(
k
β’
|
G
|
/
|
C
i
|
)
a
i
β’
k
.
With this information, we can now define the cycle index for
G
β
S
n
.
For convenience, we will denote the type of
g
β
G
β
S
n
as
T
β’
(
g
)
=
(
T
β’
(
g
)
i
β’
j
)
which is an
s
Γ
n
matrix.
We define the cycle index of
G
β
S
n
as
Z
n
β’
(
t
11
,
t
12
,
β¦
,
t
1
β’
n
,
β¦
,
t
s
β’
1
,
t
s
β’
2
,
β¦
,
t
s
β’
n
;
G
β
S
n
)
=
1
|
G
|
n
β’
n
!
β’
β
g
β
G
(
β
i
,
j
t
i
β’
j
T
β’
(
g
)
i
β’
j
)
.
The cycle index of
G
β
S
n
is a polynomial in the variables
t
i
β’
j
for
1
β€
i
β€
s
,
1
β€
j
β€
n
.
Observe that, given an
s
Γ
n
matrix
(
c
i
β’
j
)
with
β
i
,
j
j
β’
c
i
β’
j
=
n
(that is, it is the type of an element in
G
β
S
n
), the coefficient of the monomial
β
i
,
j
t
i
β’
j
c
i
β’
j
is equal to
|
Cl
β‘
(
g
)
|
|
G
|
n
β’
n
!
, where π is such that
T
β’
(
g
)
=
(
c
i
β’
j
)
and
Cl
β‘
(
g
)
denotes the conjugacy class of π in
G
β
S
n
.
Thus, the coefficient of a monomial in
Z
n
is precisely the probability that a randomly chosen element of
G
β
S
n
belongs to that conjugacy class.
The cycle index generating function of
G
β
S
n
is
1
+
β
n
=
1
β
Z
n
β’
u
n
=
1
+
β
n
=
1
β
1
|
G
|
n
β’
n
!
β’
[
β
g
β
G
(
β
i
,
j
t
i
β’
j
T
β’
(
g
)
i
β’
j
)
]
β’
u
n
.
We see some examples of cycle index in
G
β
S
n
.
Example 3.3
Let
G
=
C
2
=
{
Β±
1
}
, the cyclic group of order 2.
Then
C
2
β
S
2
is a group of order 8 which turns out to be the Dihedral group
D
8
of order 8.
Let
C
1
=
{
1
}
,
C
2
=
{
-
1
}
be the two conjugacy classes of
C
2
.
The conjugacy classes of
C
2
β
S
2
are parametrized by
2
Γ
2
matrices
(
a
i
β’
k
)
with
a
i
β’
k
β
Z
β₯
0
and
β
i
,
k
k
β’
a
i
β’
k
=
2
.
These matrices are
[
2
0
0
0
]
,
[
0
0
2
0
]
,
[
1
0
1
0
]
,
[
0
1
0
0
]
,
[
0
0
0
1
]
.
Thus, there are exactly 5 conjugacy classes in
C
2
β
S
2
.
Observe that the coefficient of each monomial in the cycle index as discussed above is the number of elements in the conjugacy class represented by that monomial divided by the size of the group.
Consider, for example, the monomial
t
11
β’
t
21
.
This monomial represents the conjugacy class parametrized by
[
1
0
1
0
]
,
whose conjugacy class size is
8
4
=
2
using Proposition 3.2, by putting
|
G
|
=
2
,
|
C
1
|
=
1
,
|
C
2
|
=
1
.
By calculating this coefficient for each monomial, we get the cycle index for
C
2
β
S
2
as follows:
Z
2
β’
(
t
11
,
t
12
,
t
21
,
t
22
;
C
2
β
S
2
)
=
1
8
β’
(
t
11
2
+
t
21
2
+
2
β’
t
11
β’
t
21
+
2
β’
t
12
+
2
β’
t
22
)
.
Example 3.4
We take a look at another example.
Consider
S
3
β
S
2
which is a group of order 72.
Let
C
1
=
{
e
}
,
C
2
=
{
(
12
)
,
(
23
)
,
(
13
)
}
,
C
3
=
{
(
123
)
,
(
132
)
}
be the three conjugacy classes in
S
3
.
The conjugacy classes in
S
3
β
S
2
are given by
3
Γ
2
matrices
(
a
i
β’
k
)
, where
a
i
β’
k
β
Z
β₯
0
and
β
i
,
k
k
β’
a
i
β’
k
=
2
.
These matrices are
[
2
0
0
0
0
0
]
,
[
0
0
2
0
0
0
]
,
[
0
0
0
0
2
0
]
,
[
1
0
1
0
0
0
]
,
[
0
0
1
0
1
0
]
,
[
1
0
0
0
1
0
]
,
[
0
1
0
0
0
0
]
,
[
0
0
0
1
0
0
]
,
[
0
0
0
0
0
1
]
.
Thus, there are nine conjugacy classes.
Once again by calculating the coefficient of each monomial using the formula for centralizers, the cycle index of
S
3
β
S
2
is given by
1
72
β’
(
t
11
2
+
9
β’
t
21
2
+
4
β’
t
31
2
+
6
β’
t
11
β’
t
21
+
12
β’
t
11
β’
t
31
+
4
β’
t
21
β’
t
31
+
6
β’
t
12
+
18
β’
t
22
+
12
β’
t
32
)
.
The following proposition expresses the cycle index generating function of
G
β
S
n
in an infinite product form, which will be useful later.
Proposition 3.5
Let πΊ be a finite group.
Let
Z
n
denote the cycle index of
G
β
S
n
.
Let
C
1
,
C
2
,
β¦
,
C
s
denote a labeling of the conjugacy classes of πΊ.
We have
1
+
β
n
=
1
β
Z
n
β’
u
n
=
β
i
=
1
s
β
j
=
1
β
exp
β‘
(
t
i
β’
j
β’
|
C
i
|
β’
u
j
|
G
|
β’
j
)
.
Proof
Observe first that the coefficient of a monomial
β
i
,
j
t
i
β’
j
a
i
β’
j
is the reciprocal of the size of the centralizer of the conjugacy class in
G
β
S
n
parametrized by the
s
Γ
n
matrix
(
a
i
β’
j
)
.
By Proposition 3.2, this is
1
β
i
,
j
a
i
β’
j
!
β’
(
j
β’
|
G
|
/
|
C
i
|
)
a
i
β’
j
=
β
i
,
j
|
C
i
|
a
i
β’
j
a
i
β’
j
!
β’
j
a
i
β’
j
β’
|
G
|
a
i
β’
j
.
Using this, it is easy to see that the coefficient of
u
n
in
β
i
=
1
s
β
j
=
1
β
exp
β‘
(
t
i
β’
j
β’
|
C
i
|
β’
u
j
|
G
|
β’
j
)
precisely yields
Z
n
which is the cycle index of
G
β
S
n
.
β
Observe that, taking
G
=
{
1
}
, the above proposition is the factorization (in infinite product form) for the cycle index generating function of
S
n
(see equation (2.2)).
4 Computing powers in
G
β
S
n
In this section, we compute the π-th powers in
G
β
S
n
, where
r
β₯
2
is prime.
We begin with some preparatory lemmas.
The proof of the following lemma is easy.
Lemma 4.1
Let
M
β₯
2
be any positive integer.
Suppose πΊ is a finite group.
Then the power map
Ο
M
:
G
β
G
defined by
g
β¦
g
M
is surjective if and only if
(
M
,
|
G
|
)
=
1
.
The following lemma is the most important for our purpose.
We set an important convention at this point.
Given an
s
Γ
n
matrix
(
a
i
β’
j
)
, we assume
a
i
β’
j
=
0
for all
i
>
s
or
j
>
n
.
We further assume, for
1
β€
i
β€
s
,
(
C
i
)
r
=
{
x
r
β£
x
β
C
i
}
,
where
C
1
,
C
2
,
β¦
,
C
s
denote the conjugacy classes of πΊ.
Observe that
(
C
i
)
r
=
C
j
for some
1
β€
j
β€
s
.
Lemma 4.2
Let
g
=
(
f
,
Ο
)
β
G
β
S
n
.
Let
T
β’
(
g
)
=
(
T
β’
(
g
)
i
β’
j
)
s
Γ
n
be the type of π.
Then the type of
g
r
is given by
T
β’
(
g
r
)
i
,
j
=
{
β
z
=
1
m
T
β’
(
g
)
y
z
,
j
+
r
β’
T
β’
(
g
)
i
,
r
β’
j
when
β’
r
β€
j
,
r
β’
T
β’
(
g
)
i
,
r
β’
j
when
β’
r
β£
j
,
where
{
y
1
,
y
2
,
β¦
,
y
m
}
β
{
1
,
2
,
β¦
,
s
}
is the complete set of indices π such that
(
C
k
)
r
=
C
i
.
Proof
Let
g
=
(
f
,
Ο
)
β
G
β
S
n
, and let
T
β’
(
g
)
=
(
T
β’
(
g
)
i
β’
j
)
s
Γ
n
denote the type of π.
Let the type of π be given by
(
c
1
,
c
2
,
β¦
,
c
n
)
.
Observe that
g
r
=
(
f
,
Ο
)
r
=
(
f
β’
f
Ο
β’
β¦
β’
f
Ο
r
-
1
,
Ο
r
)
.
Let
h
=
f
β’
f
Ο
β’
β¦
β’
f
Ο
r
-
1
β
G
.
To find
T
β’
(
g
r
)
, the type of
g
r
, we need to compute the cycle products corresponding to each cycle in
Ο
r
.
Let us fix
1
β€
i
β€
s
,
1
β€
j
β€
n
.
By Lemma 2.1, we have to consider two cases.
For the first case, let us assume that
r
β£
j
.
Then we know by Lemma 2.1 that if
(
d
1
,
d
2
,
β¦
,
d
n
)
is the type of
Ο
r
, then
d
j
=
r
β’
c
r
β’
j
.
In other words, a π-cycle in
Ο
r
can only be obtained by raising an
r
β’
j
-cycle to the power of π.
Let us consider an
r
β’
j
-cycle in π of the form
Ο
=
(
l
1
,
l
2
,
β¦
,
l
r
,
l
r
+
1
,
β¦
,
l
2
β’
r
,
β¦
,
l
(
j
-
1
)
β’
r
+
1
,
β¦
,
l
r
β’
j
)
such that the cycle product of π with respect to π is
x
=
f
β’
(
l
1
)
β’
f
β’
(
Ο
-
1
β’
(
l
1
)
)
β’
β¦
β’
f
β’
(
Ο
-
(
r
β’
j
-
1
)
β’
(
l
1
)
)
β
C
i
.
Now,
Ο
r
=
(
l
1
,
l
r
+
1
,
β¦
,
l
(
j
-
1
)
β’
r
+
1
)
β’
(
l
2
,
l
r
+
2
,
β¦
,
l
(
j
-
1
)
β’
r
+
2
)
β’
β¦
β’
(
l
r
,
l
2
β’
r
,
β¦
,
l
j
β’
r
)
.
Thus,
Ο
r
=
Ο
1
β’
Ο
2
β’
β¦
β’
Ο
r
, where
Ο
t
=
(
l
t
,
l
r
+
t
,
β¦
,
l
(
j
-
1
)
β’
r
+
t
)
.
For
1
β€
t
β€
r
, let us find the cycle product of
Ο
t
with respect to β, which is
y
=
h
β’
(
l
t
)
β’
h
β’
(
Ο
t
-
1
β’
(
l
t
)
)
β’
β¦
β’
h
β’
(
Ο
t
-
(
r
-
1
)
β’
(
l
t
)
)
=
f
β’
(
l
t
)
β’
f
β’
(
Ο
-
1
β’
(
l
t
)
)
β’
β¦
β’
f
β’
(
Ο
-
(
r
-
1
)
β’
(
l
t
)
)
β’
f
β’
(
Ο
t
-
1
β’
(
l
t
)
)
β’
f
β’
(
Ο
-
1
β’
(
Ο
t
-
1
β’
(
l
t
)
)
)
β’
β¦
f
β’
(
Ο
-
(
r
-
1
)
β’
(
Ο
t
-
1
β’
(
l
t
)
)
)
β’
β¦
β’
f
β’
(
Ο
t
-
(
r
-
1
)
β’
(
l
t
)
)
β’
f
β’
(
Ο
-
1
β’
(
Ο
t
-
(
r
-
1
)
β’
(
l
t
)
)
)
β’
β¦
f
β’
(
Ο
-
(
r
-
1
)
β’
(
Ο
t
-
(
r
-
1
)
β’
(
l
t
)
)
)
=
f
β’
(
l
t
)
β’
f
β’
(
Ο
-
1
β’
(
l
t
)
)
β’
β¦
β’
f
β’
(
Ο
-
(
r
-
1
)
β’
(
l
t
)
)
β’
f
β’
(
Ο
-
r
β’
(
l
t
)
)
β’
f
β’
(
Ο
-
(
r
+
1
)
β’
(
l
t
)
)
β’
β¦
f
β’
(
Ο
-
(
2
β’
r
-
1
)
β’
(
l
t
)
)
β’
β¦
β’
f
β’
(
Ο
-
(
j
-
1
)
β’
r
β’
(
l
t
)
)
β’
β¦
β’
f
β’
(
Ο
-
(
r
β’
j
-
1
)
β’
(
l
t
)
)
.
It is clear that π₯ is conjugate to π¦ in πΊ, and hence
y
β
C
i
.
We know
T
β’
(
g
)
i
,
r
β’
j
is the number of
r
β’
j
-cycles in π whose cycle product belongs to
C
i
.
One
r
β’
j
-cycle in π contributes π separate π-cycles in
Ο
r
, and by the above computation, the cycle product of each such π-cycle belongs to
C
i
if the cycle product of the former belongs to
C
i
.
We therefore conclude that, in the case when
r
β£
j
,
T
β’
(
g
r
)
i
,
j
=
r
β’
T
β’
(
g
)
i
,
r
β’
j
.
Let us now move on to the second case in which
r
β€
j
.
Again by Lemma 2.1,
d
j
=
c
j
+
r
β’
c
r
β’
j
.
Thus, a π-cycle in
Ο
r
is obtained from
r
β’
j
-cycles in π just as above, or from a π-cycle in π.
We need to address the case of the latter possibility since the former one has already been taken care of in the previous case.
Consider a π-cycle in π, say
Ο
=
(
l
1
,
l
2
,
β¦
,
l
j
)
.
The cycle product of π with respect to π is
x
=
f
β’
(
l
1
)
β’
f
β’
(
Ο
-
1
β’
(
l
1
)
)
β’
β¦
β’
f
β’
(
Ο
-
(
j
-
1
)
β’
(
l
1
)
)
.
Since
Ο
j
=
id
, the cycle product of
Ο
r
with respect to β is
y
=
h
β’
(
l
1
)
β’
h
β’
(
Ο
-
r
β’
(
l
1
)
)
β’
β¦
β’
h
β’
(
Ο
-
r
β’
(
j
-
1
)
β’
(
l
1
)
)
=
f
β’
(
l
1
)
β’
f
β’
(
Ο
-
1
β’
(
l
1
)
)
β’
β¦
β’
f
β’
(
Ο
-
(
r
-
1
)
β’
(
l
1
)
)
β’
f
β’
(
Ο
-
r
β’
(
l
1
)
)
β’
f
β’
(
Ο
-
1
β’
(
Ο
-
r
β’
(
l
1
)
)
)
β’
β¦
f
β’
(
Ο
-
(
r
-
1
)
β’
(
Ο
-
r
β’
(
l
1
)
)
)
β’
β¦
β’
f
β’
(
Ο
-
r
β’
(
j
-
1
)
β’
(
l
1
)
)
β’
f
β’
(
Ο
-
1
β’
(
Ο
-
r
β’
(
j
-
1
)
β’
(
l
1
)
)
)
β’
β¦
f
β’
(
Ο
-
(
r
-
1
)
β’
(
Ο
-
r
β’
(
j
-
1
)
β’
(
l
1
)
)
)
=
f
β’
(
l
1
)
β’
f
β’
(
Ο
-
1
β’
(
l
1
)
)
β’
β¦
β’
f
β’
(
Ο
-
(
r
-
1
)
β’
(
l
1
)
)
β’
f
β’
(
Ο
-
r
β’
(
l
1
)
)
β’
f
β’
(
Ο
-
(
r
+
1
)
β’
(
l
1
)
)
β’
β¦
f
β’
(
Ο
-
(
2
β’
r
-
1
)
β’
(
l
1
)
)
β’
β¦
β’
f
β’
(
Ο
-
r
β’
(
j
-
1
)
β’
(
l
1
)
)
β’
f
β’
(
Ο
-
(
r
β’
(
j
-
1
)
+
1
)
β’
(
l
1
)
)
β’
β¦
f
β’
(
Ο
-
(
r
β’
(
j
-
1
)
+
(
r
-
1
)
)
β’
(
l
1
)
)
=
f
β’
(
l
1
)
β’
f
β’
(
Ο
-
1
β’
(
l
1
)
)
β’
β¦
β’
f
β’
(
Ο
-
(
j
-
1
)
β’
(
l
1
)
)
β’
f
β’
(
Ο
-
j
β’
(
l
1
)
)
β’
f
β’
(
Ο
-
(
j
+
1
)
β’
(
l
1
)
)
β’
β¦
f
β’
(
Ο
-
(
2
β’
j
-
1
)
β’
(
l
1
)
)
β’
β¦
β’
f
β’
(
Ο
-
(
r
-
1
)
β’
j
β’
(
l
1
)
)
β’
β¦
β’
f
β’
(
Ο
-
(
(
r
-
1
)
β’
j
+
j
-
1
)
β’
(
l
1
)
)
=
f
β’
(
l
1
)
β’
f
β’
(
Ο
-
1
β’
(
l
1
)
)
β’
β¦
β’
f
β’
(
Ο
-
(
j
-
1
)
β’
(
l
1
)
)
β’
f
β’
(
l
1
)
β’
f
β’
(
Ο
-
1
β’
(
l
1
)
)
β’
β¦
f
β’
(
Ο
-
(
j
-
1
)
β’
(
l
1
)
)
β’
β¦
β’
f
β’
(
l
1
)
β’
f
β’
(
Ο
-
1
β’
(
l
1
)
)
β’
β¦
β’
f
β’
(
Ο
-
(
j
-
1
)
β’
(
l
1
)
)
=
x
.
x
β’
β¦
β’
x
β
r
β’
times
=
x
r
.
Thus, we see that the cycle product
y
β
(
C
i
)
r
.
Since, for
1
β€
k
β€
s
,
T
β’
(
g
)
k
,
j
is the number of π-cycles each of whose cycle product belongs to
C
k
, we can conclude that if
{
y
1
,
y
2
,
β¦
,
y
m
}
β
{
1
,
2
,
β¦
,
s
}
is the complete set of indices π such that
(
C
k
)
r
=
C
i
, then
T
β’
(
g
r
)
i
,
j
has a contribution of
β
z
=
1
m
T
β’
(
g
)
y
z
,
j
coming as cycle products of π-cycles in
Ο
r
.
Therefore, we can finally conclude that, when
r
β€
j
,
T
β’
(
g
r
)
i
,
j
=
β
z
=
1
m
T
β’
(
g
)
y
z
,
j
+
r
β’
T
β’
(
g
)
i
,
r
β’
j
.
This completes the proof.
β
We can now characterize the conjugacy classes in
G
β
S
n
that are π-th powers entirely in terms of the associated combinatorial data.
Proposition 4.3
Let πΊ be a finite group which has π conjugacy classes. Suppose that, of those π conjugacy classes, exactly π are not π-th powers in πΊ.
Let
C
1
,
C
2
,
β¦
,
C
s
be a labeling of the conjugacy classes of πΊ such that
C
1
,
C
2
,
β¦
,
C
d
are not π-th powers.
Let
g
β
G
β
S
n
, and let
T
β’
(
g
)
s
Γ
n
=
(
T
β’
(
g
)
i
,
j
)
denote the type of π.
Then π is an π-th power in
G
β
S
n
if and only if
r
β£
T
β’
(
g
)
i
,
j
whenever
r
β£
j
or
1
β€
i
β€
d
.
Proof
Assume that
g
β
G
β
S
n
is an π-th power, that is, there exists
h
β
G
β
S
n
such that
h
r
=
g
.
Let
T
β’
(
h
)
denote the type of β.
The type of
h
r
is given by Lemma 4.2.
Since
T
β’
(
h
r
)
=
T
β’
(
g
)
, we at once conclude that
r
β£
T
β’
(
g
)
i
,
j
whenever
r
β£
j
.
Since the conjugacy classes of πΊ are labeled as
C
1
,
C
2
,
β¦
,
C
s
such that
C
1
,
C
2
,
β¦
,
C
d
are not π-th powers, we conclude once again by Lemma 4.2 that, for
1
β€
i
β€
d
,
T
β’
(
g
)
i
,
j
=
r
β’
T
β’
(
h
)
i
,
r
β’
j
, thus showing that
r
β£
T
β’
(
g
)
i
,
j
.
Conversely, let us assume that
g
β
G
β
S
n
is such that
r
β£
T
β’
(
g
)
i
β’
j
whenever
1
β€
i
β€
d
or
r
β£
j
.
We need to show that there exists
h
β
G
β
S
n
such that
h
r
=
g
.
For
d
+
1
β€
i
β€
s
, let
C
i
1
,
C
i
2
,
β¦
,
C
i
u
be the complete set of classes such that
(
C
i
m
)
r
=
C
i
for all
1
β€
m
β€
u
.
Moreover, we choose the indices
i
m
in increasing order, that is,
i
1
<
i
2
<
β―
<
i
u
.
Clearly, for two distinct
e
,
f
β
{
d
+
1
,
β¦
,
s
}
, the indices
e
1
,
e
2
,
β¦
,
e
v
and the indices
f
1
,
f
2
,
β¦
,
f
v
β²
are disjoint.
Consider the
s
Γ
n
matrix
T
β²
defined as follows: when
r
β€
j
and
d
+
1
β€
i
β€
s
,
T
i
1
,
j
β²
=
T
β’
(
g
)
i
,
j
β
and
β
T
i
m
,
j
β²
=
0
,
where
β’
m
β
{
2
,
3
,
β¦
,
u
}
,
and
T
i
,
r
β’
j
β²
=
{
T
β’
(
g
)
i
,
j
r
if
β’
1
β€
i
β€
d
,
T
β’
(
g
)
i
,
j
r
if
β’
d
+
1
β€
i
β€
s
β’
and
β’
r
β£
j
,
0
if
β’
d
+
1
β€
i
β€
s
β’
and
β’
r
β€
j
.
In the above definition, we once again use the convention that if the indices
i
,
j
exceed the bounds
1
β€
i
β€
s
,
1
β€
j
β€
n
, we do not consider those entries.
It is also evident that, when
r
β€
j
, all the entries of
T
i
β’
j
β²
are well-defined.
Observe that
β
i
,
j
j
β’
T
i
,
j
β²
=
n
.
Indeed,
(4.1)
β
i
,
j
j
β’
T
i
,
j
β²
=
β
i
,
r
β£
j
r
β’
j
β’
T
i
,
r
β’
j
β²
+
β
i
,
r
β€
j
j
β’
T
i
,
j
β²
+
β
i
,
r
β€
j
r
β’
j
β’
T
i
,
r
β’
j
β²
.
Now, it is clear from the definition of
T
β²
that
β
i
,
r
β£
j
r
β’
j
β’
T
i
,
r
β’
j
β²
=
β
i
,
r
β£
j
j
β’
T
β’
(
g
)
i
,
j
and
β
i
,
r
β€
j
j
β’
T
i
,
j
β²
+
β
i
,
r
β€
j
r
β’
j
β’
T
i
,
r
β’
j
β²
=
β
i
,
r
β€
j
j
β’
T
i
,
j
β²
+
β
1
β€
i
β€
d
r
β€
j
r
β’
j
β’
T
i
,
r
β’
j
β²
=
β
i
,
r
β€
j
j
β’
T
i
,
j
β²
+
β
1
β€
i
β€
d
r
β€
j
j
β’
T
β’
(
g
)
i
,
j
=
β
d
+
1
β€
i
β€
s
r
β€
j
j
β’
T
β’
(
g
)
i
,
j
+
β
1
β€
i
β€
d
r
β€
j
j
β’
T
β’
(
g
)
i
,
j
=
β
i
,
r
β€
j
j
β’
T
β’
(
g
)
i
,
j
.
Thus, from equation (4.1), we get
β
i
,
j
j
β’
T
i
,
j
β²
=
β
i
,
r
β£
j
j
β’
T
β’
(
g
)
i
,
j
+
β
i
,
r
β€
j
j
β’
T
β’
(
g
)
i
,
j
=
β
i
,
j
j
β’
T
β’
(
g
)
i
,
j
=
n
.
Therefore, there exists
h
β
G
β
S
n
such that
T
β’
(
h
)
=
T
β²
, where
T
β’
(
h
)
is the type of β.
Now, by Lemma 4.2, when
r
β£
j
,
T
β’
(
h
r
)
i
,
j
=
r
β’
T
β’
(
h
)
i
,
r
β’
j
=
T
β’
(
g
)
i
,
j
.
When
r
β€
j
and
1
β€
i
β€
d
, we have
T
β’
(
h
r
)
i
,
j
=
r
β’
T
β’
(
h
)
i
,
r
β’
j
=
T
β’
(
g
)
i
,
j
.
When
r
β€
j
and
d
+
1
β€
i
β€
s
, once again by Lemma 4.2 and the fact that
i
1
,
β¦
,
i
u
are the complete set of indices such that
(
C
i
m
)
r
=
C
i
, we get
T
β’
(
h
r
)
i
,
j
=
β
m
=
1
u
T
β’
(
h
)
i
m
,
j
+
r
β’
T
β’
(
h
)
i
,
r
β’
j
=
T
β’
(
g
)
i
,
j
.
Thus,
T
β’
(
h
r
)
=
T
β’
(
g
)
, which implies that
h
r
is conjugate to π.
Therefore, there exists
x
β
G
β
S
n
such that
x
β’
h
r
β’
x
-
1
=
g
implies
(
x
β’
h
β’
x
-
1
)
r
=
g
.
This completes the proof.
β
Corollary 4.4
Let πΊ be a finite group with
r
β€
|
G
|
.
Let
g
β
G
β
S
n
, and denote by
T
β’
(
g
)
s
Γ
n
=
(
T
β’
(
g
)
i
β’
j
)
the type of π.
Then π is an π-th power in
G
β
S
n
if and only if
r
β£
T
β’
(
g
)
i
β’
j
whenever
r
β£
j
.
Proof
Since, by Lemma 4.1, all conjugacy classes in πΊ are π-th powers, we get the corollary by putting
d
=
0
in Proposition 4.3 above.
β
We end this section with some examples.
Example 4.5
Let us consider
G
=
C
3
, which is the cyclic group of order 3.
Consider
C
3
β
S
3
which is a group of order 162.
By Proposition 3.1, conjugacy classes in
C
3
β
S
3
are in one-to-one correspondence with
3
Γ
3
matrices
(
a
i
β’
j
)
with non-negative integer entries and satisfying
β
i
,
j
j
β’
a
i
β’
j
=
3
.
There are 22 conjugacy classes in
C
3
β
S
3
.
Let us list the 22 matrices that parametrize the conjugacy classes in
C
3
β
S
3
.
x
1
=
(
3
0
0
0
0
0
0
0
0
)
,
x
2
=
(
0
0
0
3
0
0
0
0
0
)
,
x
3
=
(
0
0
0
0
0
0
3
0
0
)
,
x
4
=
(
2
0
0
1
0
0
0
0
0
)
,
x
5
=
(
0
0
0
2
0
0
1
0
0
)
,
x
6
=
(
2
0
0
0
0
0
1
0
0
)
,
x
7
=
(
1
0
0
2
0
0
0
0
0
)
,
x
8
=
(
0
0
0
1
0
0
2
0
0
)
,
x
9
=
(
1
0
0
0
0
0
2
0
0
)
,
x
10
=
(
1
0
0
1
0
0
1
0
0
)
,
x
11
=
(
1
1
0
0
0
0
0
0
0
)
,
x
12
=
(
1
0
0
0
1
0
0
0
0
)
,
x
13
=
(
1
0
0
0
0
0
0
1
0
)
,
x
14
=
(
0
1
0
1
0
0
0
0
0
)
,
x
15
=
(
0
0
0
1
1
0
0
0
0
)
,
x
16
=
(
0
0
0
1
0
0
0
1
0
)
,
x
17
=
(
0
1
0
0
0
0
1
0
0
)
,
x
18
=
(
0
0
0
0
1
0
1
0
0
)
,
x
19
=
(
0
0
0
0
0
0
1
1
0
)
,
x
20
=
(
0
0
1
0
0
0
0
0
0
)
,
x
21
=
(
0
0
0
0
0
1
0
0
0
)
,
x
22
=
(
0
0
0
0
0
0
0
0
1
)
.
With this data, let us compute which conjugacy classes are squares (that is,
r
=
2
).
Observe that
|
C
3
|
=
3
, which is odd; thus, by Corollary 4.4, we conclude that the conjugacy classes which are squares are parametrized by the matrices
(
a
i
β’
j
)
1
β€
i
,
j
β€
3
with
a
i
β’
j
being even whenever π is even.
Thus, conjugacy classes from
x
1
to
x
10
and
x
20
to
x
22
are square conjugacy classes.
Thus, 13 out of the 22 conjugacy classes in
C
3
β
S
3
are squares.
A direct computation shows that exactly 81 elements are squares in
C
3
β
S
3
.
Example 4.6
Consider the wreath product
S
3
β
S
3
, which is a group of order 1296.
Let
C
1
=
{
(
12
)
,
(
23
)
,
(
13
)
}
,
C
2
=
{
e
}
,
C
3
=
{
(
123
)
,
(
132
)
}
be a labeling of the conjugacy classes in
S
3
.
Observe that
C
1
is the only non-square conjugacy class in
S
3
.
Observe again by using Proposition 3.1 the conjugacy classes in
S
3
β
S
3
are in one-to-one correspondence with
3
Γ
3
matrices
(
a
i
β’
j
)
with non-negative integer entries and
β
i
,
j
j
β’
a
i
β’
j
=
3
.
This is the same set of matrices as described in the previous example in case
C
3
β
S
3
.
Thus, the matrices
x
1
to
x
22
give the conjugacy class types of
S
3
β
S
3
once again.
We compute the squares (that is,
r
=
2
) in
S
3
β
S
3
.
Using Proposition 4.3, we conclude that the conjugacy classes in
S
3
β
S
3
that are squares are parametrized by the
3
Γ
3
matrices
(
a
i
β’
j
)
with the property that
a
i
β’
j
is even whenever either
j
=
2
or
i
=
1
.
We conclude that conjugacy classes parametrized by matrices
x
2
,
x
3
,
x
4
,
x
5
,
x
6
,
x
8
,
x
21
,
x
22
satisfy the required properties and hence are square conjugacy classes.
Therefore, in
S
3
β
S
3
, 8 conjugacy classes are squares out of the possible 22 classes.
Once again, a direct computation shows that 324 elements out of the total 1296 are squares.
5 Generating functions for the powers in
G
β
S
n
In this section, we prove one of our main results using generating functions.
Recall that, for the group
G
β
S
n
, we have
P
r
β’
(
G
β
S
n
)
=
|
Ο
r
β’
(
G
β
S
n
)
|
|
G
|
n
β’
n
!
,
where
Ο
r
β’
(
G
β
S
n
)
is the set of all π-th powers in
G
β
S
n
.
Let
Ξ±
i
=
|
C
i
|
|
G
|
for
1
β€
i
β€
s
, where
C
1
,
C
2
,
β¦
,
C
s
denote the conjugacy classes of πΊ.
For
j
β₯
1
, define the function
Ο
j
β’
(
u
)
as
Ο
j
β’
(
u
)
=
exp
β‘
(
u
j
j
)
.
Let
Ο
β
1
be an π-th root of unity in β.
We define, for
j
β₯
1
and
1
β€
i
β€
s
,
Ο
j
,
Ξ±
i
β’
(
u
)
=
1
r
β’
β
k
=
0
r
-
1
Ο
j
β’
(
u
)
Ξ±
i
β’
Ο
k
.
The following proposition gives the generating function for
P
r
β’
(
G
β
S
n
)
.
Proposition 5.1
Let πΊ be a finite group and
r
β₯
2
a prime.
Let
C
1
,
C
2
,
β¦
,
C
s
be a labeling of the conjugacy classes of πΊ in such a way that
C
1
,
C
2
,
β¦
,
C
d
are those conjugacy classes which are not π-th powers in πΊ for some
0
β€
d
<
s
.
Then
1
+
β
n
=
1
β
P
r
β’
(
G
β
S
n
)
β’
u
n
=
β
i
=
1
d
[
β
j
=
1
β
Ο
j
,
Ξ±
i
β’
(
u
)
]
β’
β
i
=
d
+
1
s
[
(
(
1
-
u
r
)
1
r
(
1
-
u
)
)
Ξ±
i
β’
β
j
=
1
β
Ο
r
β’
j
,
Ξ±
i
β’
(
u
)
]
.
Proof
We will make use of the cycle index generating function of
G
β
S
n
in Proposition 3.5 and the characterization of π-th powers given by Proposition 4.3.
By Proposition 3.5, we have
(5.1)
1
+
β
n
=
1
β
Z
n
β’
u
n
=
β
i
=
1
s
β
j
=
1
β
exp
β‘
(
t
i
β’
j
β’
Ξ±
i
β’
u
j
j
)
.
By Proposition 4.3, we know that if π is an π-th power in
G
β
S
n
, then each entry in the first π rows of
T
β’
(
g
)
is divisible by π, where
T
β’
(
g
)
=
type
β‘
(
g
)
.
Therefore, to include the conjugacy classes in
G
β
S
n
which are π-th powers in
G
β
S
n
for
1
β€
i
β€
d
,
j
β₯
1
, we put
t
i
β’
j
=
1
in equation (5.1).
Now, only the coefficients of
u
m
β’
r
need to be included in the formal power series expansion of
exp
β‘
(
Ξ±
i
β’
u
j
j
)
for each
m
β₯
1
.
This is achieved when we replace
exp
β‘
(
Ξ±
i
β’
u
j
j
)
by
Ο
j
,
Ξ±
i
β’
(
u
)
.
For the rows from
d
+
1
to π in
T
β’
(
g
)
, the entries in the π-th column are multiples of π whenever
r
β£
j
.
Thus, for
d
+
1
β€
i
β€
s
and for each
j
β₯
1
, we put
t
i
β’
j
=
1
whenever
r
β€
j
.
Whenever
r
β£
j
, we put
t
i
β’
j
=
1
as well as replace
exp
β‘
(
Ξ±
i
β’
u
j
j
)
by
Ο
j
,
Ξ±
i
β’
(
u
)
in equation (5.1) by the same argument as in the previous case.
Thus, plugging the above constraints in equation (5.1), we get
(5.2)
1
+
β
n
=
1
β
P
r
β’
(
G
β
S
n
)
β’
u
n
=
β
i
=
1
d
[
β
j
=
1
β
Ο
j
,
Ξ±
i
β’
(
u
)
]
β’
β
i
=
d
+
1
s
[
β
j
=
1
r
β€
j
β
exp
β‘
(
Ξ±
i
β’
u
j
j
)
]
β’
[
β
j
=
1
r
β£
j
β
Ο
j
,
Ξ±
i
β’
(
u
)
]
.
Now,
β
j
=
1
r
β€
j
β
exp
β‘
(
u
j
j
)
=
β
j
=
1
β
exp
β‘
(
u
j
j
)
β
j
=
1
,
r
β£
j
β
exp
β‘
(
u
j
j
)
=
1
(
1
-
u
)
β’
β
j
=
1
,
r
β£
j
β
exp
β‘
(
u
j
j
)
=
1
(
1
-
u
)
β’
β
j
=
1
β
exp
β‘
(
u
r
β’
j
r
β’
j
)
=
1
(
1
-
u
)
β’
exp
β‘
(
-
1
r
β’
log
β‘
(
1
-
u
r
)
)
=
(
1
-
u
r
)
1
r
1
-
u
.
Thus, the final result follows from the above computation and equation (5.2).β
Corollary 5.2
Let
r
β₯
2
be a prime and πΊ a finite group with
r
β€
|
G
|
.
Assume that πΊ has π conjugacy classes.
Then
1
+
β
n
=
1
β
P
r
β’
(
G
β
S
n
)
β’
u
n
=
(
(
1
-
u
r
)
1
r
(
1
-
u
)
)
β’
β
i
=
1
s
β
j
=
1
β
Ο
r
β’
j
,
Ξ±
i
β’
(
u
)
.
Proof
Since
β
i
=
1
s
Ξ±
i
=
1
, the proof follows from the above proposition where we take
d
=
0
due to Corollary 4.4.
β
Example 5.3
Let
G
=
C
3
be the cyclic group of order 3.
Let
r
=
2
.
By the above corollary,
1
+
β
n
=
1
β
P
2
β’
(
C
3
β
S
n
)
β’
u
n
=
(
1
+
u
1
-
u
)
1
2
β’
β
j
β₯
1
cosh
3
β‘
(
u
2
β’
j
6
β’
j
)
,
where
cosh
n
β‘
(
x
)
=
(
cosh
β‘
x
)
n
.
Example 5.4
The Weyl group of type
B
n
(or
C
n
) for
n
β₯
2
is the wreath product
C
2
β
S
n
.
Using Corollary 5.2, the generating function for the proportion of squares in these groups can be given as follows:
1
+
β
n
=
1
β
P
2
β’
(
B
n
)
β’
u
n
=
1
+
β
n
=
1
β
P
2
β’
(
C
n
)
β’
u
n
=
(
1
+
u
1
-
u
)
1
4
β’
β
j
β₯
1
cosh
β‘
(
u
j
2
β’
j
)
β’
cosh
β‘
(
u
2
β’
j
4
β’
j
)
,
where we assume
B
1
=
C
1
=
C
2
and
P
2
β’
(
B
n
)
(resp.
P
2
β’
(
C
n
)
) denotes the proportion of squares in
B
n
(resp.
C
n
).
In particular, if we take
G
=
{
1
}
and
r
=
2
in Corollary 5.2, we get back the generating function for the proportion of squares in
S
n
(see equation (2.3)).
Now, we proceed towards proving our main theorem.
The following lemma which is a generalization of [4, Lemma 2] will be needed.
Lemma 5.5
Let π be a prime, and let
Ο
β
1
be an π-th root of unity in β.
Let
f
β’
(
x
)
be a formal power series such that
f
β’
(
x
)
=
f
β’
(
Ο
β’
x
)
.
Suppose
g
β’
(
x
)
=
(
1
+
x
+
β―
+
x
r
-
1
)
β’
f
β’
(
x
)
=
β
n
=
0
β
a
n
β’
x
n
.
Then
a
k
+
1
=
a
k
for
k
β’
-
1
β’
(
mod
β’
r
)
.
Proof
Let
f
β’
(
x
)
=
β
n
=
0
β
b
n
β’
x
n
.
Since
f
β’
(
x
)
=
f
β’
(
Ο
β’
x
)
, we have
b
i
=
0
for
r
β€
i
.
Equating the coefficient of
x
n
on both sides of
(
1
+
x
+
β―
+
x
r
-
1
)
β’
f
β’
(
x
)
=
β
n
=
0
β
a
n
β’
x
n
,
we get
a
k
+
1
=
a
k
for
k
β’
-
1
β’
(
mod
β’
r
)
.
β
Theorem 5.6
Let πΊ be a finite group and
r
β₯
2
a prime with
(
r
,
|
G
|
)
=
1
.
Then
P
r
β’
(
G
β
S
n
+
1
)
=
P
r
β’
(
G
β
S
n
)
for every
n
β
N
with
n
β’
-
1
β’
(
mod
β’
r
)
.
In other words,
|
Ο
r
β’
(
G
β
S
n
+
1
)
|
=
|
G
|
β’
(
n
+
1
)
β’
|
Ο
r
β’
(
G
β
S
n
)
|
for every
n
β’
-
1
β’
(
mod
β’
r
)
.
Proof
We will show that
g
β’
(
u
)
=
1
+
β
n
=
1
β
P
r
β’
(
G
β
S
n
)
β’
u
n
is such that
g
β’
(
u
)
=
(
1
+
u
+
β―
+
u
r
-
1
)
β’
f
β’
(
u
)
,
for a function
f
β’
(
u
)
with
f
β’
(
u
)
=
f
β’
(
Ο
β’
u
)
, where
Ο
β
1
is an π-th root of unity.
Then, by Lemma 5.5, the result holds.
By Corollary 5.2, we have
(5.3)
g
β’
(
u
)
=
(
(
1
-
u
r
)
1
r
(
1
-
u
)
)
β’
β
i
=
1
s
β
j
=
1
β
Ο
r
β’
j
,
Ξ±
i
β’
(
u
)
.
We have
(
1
-
u
r
)
1
r
(
1
-
u
)
=
(
1
-
u
r
1
-
u
)
β’
(
1
-
u
r
)
1
-
r
r
=
(
1
+
u
+
β―
+
u
r
-
1
)
β’
(
1
-
u
r
)
1
-
r
r
.
Thus, substituting the above in equation 5.3, we get
g
β’
(
u
)
=
(
1
+
u
+
β―
+
u
r
-
1
)
β’
(
1
-
u
r
)
1
-
r
r
β’
β
i
=
1
s
β
j
=
1
β
Ο
r
β’
j
,
Ξ±
i
β’
(
u
)
.
Take
f
β’
(
u
)
=
(
1
-
u
r
)
1
-
r
r
β’
β
i
=
1
s
β
j
=
1
β
Ο
r
β’
j
,
Ξ±
i
β’
(
u
)
.
Then
g
β’
(
u
)
=
(
1
+
u
+
β―
+
u
r
-
1
)
β’
f
β’
(
u
)
.
By the discussion at the beginning of the proof, it remains to prove
f
β’
(
u
)
=
f
β’
(
Ο
β’
u
)
.
Since
Ο
r
=
1
, it is clear that
Ο
j
,
Ξ±
i
β’
(
Ο
β’
u
)
=
Ο
j
,
Ξ±
i
β’
(
u
)
for all
j
β₯
1
and
1
β€
i
β€
s
.
Finally,
(
1
-
(
Ο
β’
u
)
r
)
1
-
r
r
=
(
1
-
u
r
)
1
-
r
r
.
This proves that
f
β’
(
u
)
=
f
β’
(
Ο
β’
u
)
, and the proof follows.
Moreover, by definition,
P
r
β’
(
G
β
S
n
+
1
)
=
P
r
β’
(
G
β
S
n
)
implies that
|
Ο
r
β’
(
G
β
S
n
+
1
)
|
=
|
G
|
β’
(
n
+
1
)
β’
|
Ο
r
β’
(
G
β
S
n
)
|
.
β
Remark 5.7
If
(
r
,
|
G
|
)
=
1
for some finite group πΊ, by Lemma 4.1, the power map
Ο
r
:
G
β
G
is surjective, and therefore, it might naturally seem that
|
Ο
r
β’
(
G
β
S
n
)
|
=
|
G
|
n
β’
|
Ο
r
β’
(
S
n
)
|
.
In such a case, the above theorem trivially follows from Theorem 2.3.
But we remark that
|
Ο
r
β’
(
G
β
S
n
)
|
=
|
G
|
n
β’
|
Ο
r
β’
(
S
n
)
|
does not hold.
As an example, consider
G
=
C
3
, the cyclic group of order 3.
Then the wreath product
C
3
β
S
4
is a group of order 1944.
Using Corollary 4.4, taking
r
=
2
, we see that
|
Ο
2
β’
(
C
3
β
S
4
)
|
=
810
.
Since
|
Ο
2
β’
(
S
4
)
|
=
12
, we have
|
Ο
2
β’
(
C
3
β
S
4
)
|
=
810
β
3
4
β’
.12
=
972
.
Finally, we give a formula to compute
C
β’
C
r
β’
(
G
β
S
n
)
, the number of conjugacy classes in
G
β
S
n
that are π-th powers.
Let
C
β’
C
β’
(
G
β
S
n
)
denote the number of conjugacy classes in
G
β
S
n
.
Let the number of conjugacy classes in πΊ be π .
Then it is known (see [7, 3.8]) that
C
β’
C
β’
(
G
β
S
n
)
=
β
n
1
,
n
2
,
β¦
,
n
s
p
β’
(
n
1
)
β’
p
β’
(
n
2
)
β’
β¦
β’
p
β’
(
n
s
)
,
where
p
β’
(
t
)
denotes the number of partitions of π‘ and the above sum runs over all ordered π -tuples
(
n
1
,
n
2
,
β¦
,
n
s
)
such that
n
i
β₯
0
for every
1
β€
i
β€
s
, and
β
i
=
1
s
n
i
=
n
.
The generating function for
C
β’
C
β’
(
G
β
S
n
)
is easy to write by virtue of the explicit formula for
C
β’
C
β’
(
G
β
S
n
)
.
Recall that the generating function for
p
β’
(
n
)
, which is the number of partitions of π, is
(5.4)
P
β’
(
u
)
=
1
+
β
n
=
1
β
p
β’
(
n
)
β’
u
n
=
β
i
=
1
β
(
1
1
-
u
i
)
.
By the formula for
C
β’
C
β’
(
G
β
S
n
)
and using the Cauchy product, we get the generating function for
C
β’
C
β’
(
G
β
S
n
)
as follows.
Proposition 5.8
We have
1
+
β
n
=
1
β
C
β’
C
β’
(
G
β
S
n
)
β’
u
n
=
β
i
=
1
β
(
1
1
-
u
i
)
s
,
where π denotes the number of conjugacy classes of πΊ.
To write the formula for
C
β’
C
r
β’
(
G
β
S
n
)
, we will need to introduce some notation.
Let
p
r
β’
(
n
)
denote the number of partitions
Ξ»
=
1
m
1
β’
2
m
2
β’
β¦
β’
n
such that
r
β£
m
i
for each
i
β₯
1
.
Let
p
r
β²
β’
(
n
)
denote the number of partitions
Ξ»
β’
n
such that
r
β£
m
r
β’
i
for each
i
β₯
1
.
We further assume that
p
β’
(
0
)
=
p
r
β’
(
0
)
=
p
r
β²
β’
(
0
)
=
1
.
Theorem 5.9
Let
r
β₯
2
be a prime and πΊ a finite group.
Suppose that, out of the π conjugacy classes of πΊ, exactly π are not π-th powers in πΊ.
Then
C
β’
C
r
β’
(
G
β
S
n
)
=
β
n
1
,
n
2
,
β¦
,
n
s
p
r
β’
(
n
1
)
β’
β¦
β’
p
r
β’
(
n
d
)
β’
p
r
β²
β’
(
n
d
+
1
)
β’
β¦
β’
p
r
β²
β’
(
n
s
)
,
where the above sum runs over all ordered π -tuples
(
n
1
,
n
2
,
β¦
,
n
s
)
such that
n
i
β₯
0
for every
1
β€
i
β€
s
and
β
i
=
1
s
n
i
=
n
.
Proof
By Proposition 4.3, we can conclude that the conjugacy classes in
G
β
S
n
that are π-th powers are in one-to-one correspondence with
s
Γ
n
matrices
(
a
i
β’
j
)
with non-negative integer entries satisfying (1)
β
i
,
j
j
β’
a
i
β’
j
=
n
, (2) entries in the first π rows are multiples of π, (3) entries in those columns indexed by multiples of π are divisible by π.
Since, for
1
β€
i
β€
s
, the entries
(
a
i
β’
j
)
1
β€
j
β€
n
in the π-th row give a partition of
n
i
=
β
j
=
1
n
j
β’
a
i
β’
j
, we can conclude that, for each of the first π rows, since all entries are multiples of π, each row can be chosen in
p
r
β’
(
n
i
)
ways.
For
d
+
1
β€
i
β€
s
, we see that a row
(
a
i
β’
j
)
1
β€
j
β€
n
once again defines a partition of
n
i
as explained above, but now this partition has the property that each part which is a multiple of π occurs with multiplicity divisible by π.
This is ensured by condition (3) of the specified combinatorial data.
Thus, for each
d
+
1
β€
i
β€
s
, each π-th row can be chosen in
p
r
β²
β’
(
n
i
)
ways.
Finally, observe that
β
i
=
1
s
n
i
=
n
by condition (1) of the combinatorial data.
Thus, by the principle of multiplication, we conclude that the number of combinatorial data that satisfies the three conditions specified above (which is equal to the number of conjugacy classes that are π-th powers) is
β
n
1
,
n
2
,
β¦
,
n
s
p
r
β’
(
n
1
)
β’
β¦
β’
p
r
β’
(
n
d
)
β’
p
r
β²
β’
(
n
d
+
1
)
β’
β¦
β’
p
r
β²
β’
(
n
s
)
,
where the above sum runs over all ordered pairs
(
n
1
,
n
2
,
β¦
,
n
s
)
satisfying
n
i
β₯
0
for every
1
β€
i
β€
s
, and
β
i
=
1
s
n
i
=
n
.
β
Example 5.10
Let us count the number of conjugacy classes that are squares in
S
3
β
S
3
.
So
r
=
2
,
n
=
3
.
Since
S
3
has a total of 3 conjugacy classes, out of which exactly two are squares, we have
s
=
3
,
d
=
1
.
We have Table 1 for
p
β’
(
n
)
,
p
2
β’
(
n
)
,
p
2
β²
β’
(
n
)
for
n
=
1
,
2
,
3
.
Table 1
p
β’
(
n
)
,
p
2
β’
(
n
)
,
p
2
β²
β’
(
n
)
for
n
=
1
,
2
,
3
.
π |
p
β’
(
n
)
|
p
2
β’
(
n
)
|
p
2
β²
β’
(
n
)
|
1 |
1 |
0 |
1 |
2 |
2 |
1 |
1 |
3 |
3 |
0 |
2 |
Using the formula in Theorem 5.9 and using Table 1, we have
C
β’
C
2
β’
(
S
3
β
S
3
)
=
p
2
β’
(
3
)
β’
p
2
β²
β’
(
0
)
β’
p
2
β²
β’
(
0
)
+
p
2
β’
(
0
)
β’
p
2
β²
β’
(
3
)
β’
p
2
β²
β’
(
0
)
+
p
2
β’
(
0
)
β’
p
2
β²
β’
(
0
)
β’
p
2
β²
β’
(
3
)
+
p
2
β’
(
2
)
β’
p
2
β²
β’
(
1
)
β’
p
2
β²
β’
(
0
)
+
p
2
β’
(
2
)
β’
p
2
β²
β’
(
0
)
β’
p
2
β²
β’
(
1
)
+
p
2
β’
(
1
)
β’
p
2
β²
β’
(
2
)
β’
p
2
β²
β’
(
0
)
+
p
2
β’
(
1
)
β’
p
2
β²
β’
(
0
)
β’
p
2
β²
β’
(
2
)
+
p
2
β’
(
0
)
β’
p
2
β²
β’
(
1
)
β’
p
2
β²
β’
(
2
)
+
p
2
β’
(
0
)
β’
p
2
β²
β’
(
2
)
β’
p
2
β²
β’
(
1
)
+
p
2
β’
(
1
)
β’
p
2
β²
β’
(
1
)
β’
p
2
β²
β’
(
1
)
=
8
,
which agrees with our direct calculation in Example 4.5.
We write the generating function for
C
β’
C
r
β’
(
G
β
S
n
)
.
The generating function for
p
r
β’
(
n
)
and
p
r
β²
β’
(
n
)
is easy to deduce.
We have
P
r
β’
(
u
)
=
1
+
β
n
=
1
β
p
r
β’
(
n
)
β’
u
n
=
β
i
=
1
β
1
1
-
u
r
β’
i
=
P
β’
(
u
r
)
.
The last equality comes from equation (5.4).
Further,
P
r
β²
β’
(
u
)
=
1
+
β
n
=
1
β
p
r
β²
β’
(
n
)
β’
u
n
=
β
i
=
1
β
1
(
1
-
u
r
β’
i
-
1
)
β’
β¦
β’
(
1
-
u
r
β’
i
-
(
r
-
1
)
)
β’
(
1
-
u
r
2
β’
i
)
.
Using the formula for
C
β’
C
r
β’
(
G
β
S
n
)
in the previous theorem and the Cauchy product of formal power series, we conclude as follows.
Proposition 5.11
Let
r
β₯
2
be a prime and πΊ a finite group.
Suppose that, out of the π conjugacy classes of πΊ, exactly π are not π-th powers.
Then
1
+
β
n
=
1
β
C
β’
C
r
β’
(
G
β
S
n
)
β’
u
n
=
P
β’
(
u
r
)
d
β’
P
r
β²
β’
(
u
)
s
-
d
.
5.1 Questions on asymptotics of the power map on
G
β
S
n
We end this section by briefly discussing some asymptotic aspects of the powers in
G
β
S
n
.
It is clear that, for
r
β₯
2
,
|
Ο
r
(
G
β
S
n
)
|
]
β€
|
G
|
n
|
Ο
r
(
S
n
)
|
βΉ
P
r
(
G
β
S
n
)
β€
|
Ο
r
β’
(
S
n
)
|
n
!
=
P
r
(
S
n
)
.
From the above inequality, it is clear that, for a fixed πΊ and prime
r
β₯
2
,
lim
n
β
β
β‘
P
r
β’
(
G
β
S
n
)
=
0
due to the Sandwich theorem and the fact that
lim
n
β
β
β‘
P
r
β’
(
S
n
)
=
0
(see [6]).
It will be interesting to find estimates for the powers in
G
β
S
n
analogous to those in
S
n
(see Section 2).
We finish this paper with an interesting question on the distribution of the values
P
r
β’
(
G
β
S
n
)
for various πΊ that we observed using some GAP calculations on the powers.
Question 5.12
Fix
n
β₯
2
.
Let
r
β₯
2
be a prime.
Suppose πΊ is a finite group with
(
r
,
|
G
|
)
=
1
.
Is it true that
P
r
β’
(
S
n
+
1
)
β€
P
r
β’
(
G
β
S
n
)
β€
P
r
β’
(
S
n
)
when
n
β‘
-
1
β’
(
mod
β’
r
)
?
This means that, for a fixed π, the values
P
r
β’
(
G
β
S
n
)
are distributed between
P
r
β’
(
S
n
+
1
)
and
P
r
β’
(
S
n
)
.
In fact, this compels us to ask another natural question.
Question 5.13
Does the following result hold: for a fixed π with
n
β‘
-
1
β’
(
mod
β’
r
)
and given
Ξ΅
>
0
, there exists
N
β
N
such that
P
r
β’
(
G
β
S
n
)
-
P
r
β’
(
S
n
+
1
)
<
Ξ΅
for all πΊ with
|
G
|
>
N
and
r
β€
|
G
|
.
This in fact says that, as the size of πΊ grows larger, the value of
P
r
β’
(
G
β
S
n
)
converges to
P
r
β’
(
S
n
+
1
)
.
The fact that every element is an π-th power in πΊ when
r
β€
|
G
|
in some sense controls the growth of
P
r
β’
(
G
β
S
n
)
, hence giving a strong indication that the above results can be answered in the positive.
Remark 5.14
The above result definitely does not hold if we remove the condition that
r
β€
|
G
|
.
Suppose
G
=
C
2
, the cyclic group of order 2.
Consider
r
=
2
.
Take
n
=
3
.
We have
P
2
β’
(
S
3
)
=
1
2
=
P
2
β’
(
S
4
)
.
But
P
2
β’
(
C
2
β
S
3
)
=
1
4
<
1
2
.
We end this paper by indicating another possible generalization of this work which might be worth considering.
Although the Weyl groups of type
A
n
,
B
n
,
C
n
are wreath products, the Weyl group of type
D
n
is not.
The Weyl group of type
D
n
is the semidirect product
C
2
n
-
1
β
S
n
.
More generally, the finite imprimitive unitary reflection group denoted by
G
β’
(
m
,
p
,
n
)
, where
p
β£
m
, is defined as follows:
consider the group
A
β’
(
m
,
p
,
n
)
=
{
(
ΞΈ
1
,
β¦
,
ΞΈ
n
)
β
ΞΌ
m
n
|
(
β
i
=
1
n
ΞΈ
i
)
m
p
=
1
}
.
Here,
ΞΌ
m
is the group of π-th roots of unity in β and
ΞΌ
m
n
is the direct product of π copies of
ΞΌ
m
.
Then
G
β’
(
m
,
p
,
n
)
is the semidirect product
A
β’
(
m
,
p
,
n
)
β
S
n
(see [10, Chapter 2] for more details).
In particular, the Weyl group of type
D
n
is
G
β’
(
2
,
2
,
n
)
.
The groups
G
β’
(
m
,
p
,
n
)
are normal subgroups of
G
β’
(
m
,
1
,
n
)
(
=
C
m
β
S
n
) of index π.
It will be interesting to determine the set of powers in these groups and deduce estimates on the proportion of powers (by a generating function approach or otherwise).
We intend to take this up as a possible future work.
Since
S
n
is
G
β’
(
1
,
1
,
n
)
, such results will also generalize the results on powers in
S
n
(see Section 2) in another direction.
We thank Prof. Michael Giudici for bringing this to our attention.