Advances in Linear Algebra & Matrix Theory
 Vol.2 No.2(2012), Article ID:20470,7 pages DOI:10.4236/alamt.2012.22002

Solving the Class Equation in an Alternating Group for Each and

Shuker Mahmood1,2, Andrew Rajah2

1Department of Mathematics, College of Science, University of Basra, Basra, Iraq

2School of Mathematical Sciences, Universiti Sains Malaysia, Penang, Malaysia

Email: shuker.alsalem@gmail.com, andy@cs.usm.my

Received March 6, 2012; revised April 1, 2012; accepted May 1, 2012

Keywords: Alternating Groups; Permutations; Conjugate Classes; Cycle Type; Frobenius Equation

ABSTRACT

The main purpose of this paper is to solve the class equation in an alternating group, (i.e. find the solutions set) and find the number of these solutions where ranges over the conjugacy class in and d is a positive integer. In this paper we solve the class equation in where, for all. is the complement set of where { of, with all parts of are different and odd}. is conjugacy class of and form class depends on the cycle type of its elements If and, then splits into the two classes of.

1. Introduction

The Frobenius equation in finite groups was introduced by G. Frobenius and then was studied by many others such as ([1-4]). Where they dealt with some types of finite groups like finite cyclic groups, finite pgroups, Wreath products of finite groups, etc. Choose any and write it as. With disjoint cycles of length and is the number of disjoint cycle factors including the 1-cycle of. Since disjoint cycles commute, we can assume that

. Therefore is a partition of n and it is call cycle type of. Let be the set of all elements with cycle type, then we can determine the conjugate class of by using cycle type of, since each pair of and in are conjugate if they have the same cycle type (see [5]). Therefore, the number of conjugacy classes of is the number of partitions of n. However, this is not necessarily true in an alternating group. Let and are two permutations in we have they are belong to the same conjugate class in (i.e.) since

that means they have the same cycle type but in fact and are not conjugate in, also let

and in we have they are belong to the same conjugate class in since

but here they are conjugate in. So from the first and second examples we consider it is not necessarily if two permutations have the same cycle type are conjugate in therefore in this work we discuss  in detail the conjugacy classes in an alternating group and we denote to conjugacy class of in by. Also we introduce some theorems to solve the class equation in where, for all.

1.1. Definition [6]

A partition is a sequence of nonnegative integers with and. The length

and the size of are defined as

and. We set for. An element of is called a partition of n.

1.2. Remark [6]

We only write the non zero components of a partition. Choose any and write it as. With disjoint cycles of length and is the number of disjoint cycle factors including the 1-cycle of. Since disjoint cycles commute, we can assume that

. Therefore is a partition of n and each is called part of.

1.3. Definition [6]

We call the partition

the cycle type of.

1.4. Definition [6]

Let be a partition of n. We define to be the set of all elements with cycle type.

1.5. Definition [6]

Let be given. We define to be the number of cycles of length m of.

1.6. Remarks

1) If, then we write.

2) The relationship between partitions and is as follows: if is given then, (see [6])

3) The cardinality of each can be found as follows: with and, (see [7]).

4) splits into two An-classes of equal order iff, and the non-zero parts of are different and odd, in every other case does not split, (see [8]).

1.7. Lemma [9]

Let p prime number and a conjugate class of symmetric group. If p does not divide a, then the solutions of are:

1), if

2) if

1.8. Lemma [9]

Let p and q be different prime numbers and a conjugate class of symmetric group. If and q does not divide a, then the solutions of are:

1), where i and j are solutions of the equation if.

2) No solution if p does not divide r.

1.9. Lemma [9]

Let p and q be different prime numbers and a conjugate class in Sn. If p does not divide a and q does not divide a, then the solutions of are, where i, j, k and l are nonnegative integers and solutions of the equation i + pj + qk + pql = r.

2. Conjugacy Class 2075  /> of An [10]

Let, where is a permutation in an alternating group. We define the conjugacy class of in by:

where {of, with all parts of different and odd}.

2.1. Remarks

1).

2), where is complement of.

3) and split into two classes of.

4) If, and, then

5) If, then for each, is conjugate to in.

2.2. Definition

Let { of the number of parts of with the property (mod 4) is odd}. Then, for each, of is defined by

,

.

2.3. Definition

Let { of the number of parts of with the property (mod 4) is even}. Then, for each, of is defined by

,

where does not conjugate to.

3. Results for Even Permutations in

3.1. Theorem

Let be the conjugacy class of in. If p is a prime number and does not divide a, , where is a class of, then the solutions of are

1) if and (a is odd or (a and r) are even)

2) if [((a and p) are odd) or (p is odd and (a and r) are even)] and

3)

if [(a is odd and p is even) or (a, p and r are even)] and [and m is even]

4)

if [(a is odd and p is even) or (a, p and r are even)] and [and m is odd]

5) if

[(a and p) are even and r is odd] and [and m is odd]

6)

if [(a and p) are even and r is odd)] and [and m is even], or

7) Does not exist, if [(a is even and (p and r) are odd].

Proof

Given that, , then by (1.7), the solutions of in are

a), if, or

b) if.

1) Assume and (a is odd or (a and r) are even), then from a), is the solution set of in. Let. If a is odd andwhere (odd) for each, then is a product of an even number similar to of transpositions for all. For any r (odd or even), is a product of = (even) number of transpositions. If a and r are even and, where (even) for each, then is a product of an odd number similar to of transpositions for all is a product of

= (even) number of transpositions , then the solution set of in An is.

2) Assume [(a and p) are odd) or (p is odd and (a and r) are even)] and then from b),

are solutions of in Sn. Let, considering that (a is odd or (a and r) are even) and for each . If a and p are odd, then, where and

is a product of an even number of transpositions for all, is a product of an even number of transpositions, and

, where (odd),

is a product of an even number of transpositions for all and is a product of an even number of transpositions. If (p is odd and (a and r) are even), then is a product of an odd number similar to Li of transpositions,

. Moreover, (even), and

is a product of an odd number similar to of transpositions for all. If k is odd, then is a product of + = (odd) + (odd) = (even) number of transpositions. If k is even, then is a product of + = (even) + (even) = (even) number of transpositions, then the solutions of in are

.

3) and 4) Assume [(a is odd and p is even) or (a, p, and r are even)] and. Then, from b), , , and

are solutions of in. Let, [a is odd or (a and r) are even]. For each, if (a is odd and p is even), where and (odd), is a product of an even number similar to of transpositions, and, where (even),

is a product of an odd number similar to of transpositions for all. If k is odd, then is a product of + = (even) + (odd) = (odd) number of transpositions. If k is even, then is a product of + = (even) + (even) = (even) number of transpositions . If (a, p, and r are even), then, where and (even),

is a product of an odd number similar to of transpositions, and, where (even),

is a product of an odd number similar to of transpositions for all. If k is odd, then is a product of + = (even) + (odd) = (odd) number of transpositions. If k is even, then is a product of + = (even) + (even) = (even) number of transpositions, then the solutions of in are

(if m is even) and

(if m is odd).

5) and 6) Assume [(a and p) are even and r is odd)] and. From b),  

are solutions of in. Let

(even),

is a product of an odd number similar to of transpositions for all, is a product of (odd) number of transposetions, also for each, where

, where (even), and, where

(even) is product of an odd number similar to of transpositions for each, and is product of an odd number similar to of transpositions for each. If k is odd, then is a product of + = (odd) + (odd) = (even) number of transpositions. If k is even, then is a product of + = (odd) + (even) = (odd) number of transpositions . Then, if [(a and p) are even and r is odd)] and, then the solutions of in are

(if m is odd), or

(if m is even).

7) Assume (a is even and (p and r) are odd). For each or , then there is no solution of in.

3.2. Remarks

Let p and q be different prime numbers and a conjugate class of symmetric group. If, and q does not divide a we defined collection of sets of conjugate classes of as following:

1) and j are non-negative and solutions of the equation

2)

3)

4)

5)

*We note that &

**,

,

,

3.3. Remarks

1) If a, p and q are odd, then for each, where we have is even.

2) If a is even, then for each, where we have (is even if) and (is odd if).

3) If p is even, then for each, where we have (is even if) and (is odd if).

4) If q is even and a, p are odd, then for each, where we have (is even if) and (is odd if).

3.4. Theorem

Let be a conjugacy class of in, and

, where is a class of. If p and q are different two prime numbers and and q does not divide a, then the solutions of in are:

1) W if and (a, p and q are odd).

2) if and (a or p is even).

3) if and (q is even & (a and p) are odd).

4) Not exist if p does not divide r.

5) Not exist if, (a or p is even) and.

6) Not exist if, (q is even  & (a and p) are odd) and.

Proof:

Since, and by (1.8) we have that the solution of in is:

a) W if.

b) Not exist if p does not divide r.

1) Assume and (a, p and q are odd). Then from a) we have W is the solution set of in. Let for each, we have is even permutation  Then the solution set in is W.

2) Assume and (a or p is even). Then from a) we have W is the solution set of in. Let for each, we have (is even permutation, if) and (is odd permutation, if). Then the solution set in is .

3) Assume and (q is even & (a and p) are odd). Then from a) we have W is the solution set of in. Let for each, we have (is even permutation, if) and (is odd permutation, if). Then the solution set in is.

4) Assume p does not divide r. Then from b) we have no solution of in no solution of in.

5) and 6) it is clear if, (a or p is even) and, then and there exists no solution in, also if, (q is even & (a and p) are odd) and, then and there exists no solution in.

3.5. Remarks

Let p and q be two different prime numbers and a conjugate class of symmetric group, p does not divide a and q does not divide a we defined a collection of sets of conjugate classes of as following:

.

.

.

.

.

.

.

.

.

.

We can denote D as the following: 

•

•

•

3.6. Remarks

1) If a, p and q are odd, then for each, where, is even.

2) If a is even, then for each, is even if and is odd if .

3) If p is even and (a and q) are odd, then for each, is even if and is odd if.

4) If q is even and (a and p) are odd, then for each, is even if and is odd if.

3.7. Theorem

Let be the conjugacy class of in, and, where is a class of, p and q are different two prime numbers. If p does not divide a and q does not divide a, then the solution of in is

1) D, if a, p and q are odd

2), where, if a is even

3), if a and q are odd, and p is even

4), if a and p are odd, and q is even

5) does not exist, ifaiseven, and

6) does not exist, if p is even, a and q are odd, and, or

7) does not exist, if q is even, a and p are odd, and.

Proof

Considering that, then by (2.2.11), D is the solution set of in.

1) Assume a, p and q are odd. Let for each is even. Then the solution set in is D if a, p and q are odd.

2) Assume a is even. Let for each is even. Then the solution set in is, if a is even.

3) Assume a and q are odd, and p is even. Let for each is even. Then the solution set in is, if a and q are odd, and p is even.

4) Assume a and p are odd, and q is even. Let for each is even. Then the solution set in is, if a and p are odd, and q is even.

5) Assume a is even and . Then there is no solution in.

6) and 7) Assume a and q are odd, p is even, and. Then there is no solution in, also if a and p are odd, q is even, and. Then there is no solution in.

3.8. The Number of the Solutions

Assume are even permutations where and and for all 1 ≤ t ≤ T and 1 ≤ j ≤ k. Then we can find the number of the solutions of class equation in, where d is a positive integer number as follow:

1) If are the solutions, then the number of solutions set is

2) If are the solutions, then the number of solutions set is

3) If are the solutions, then the number of solutions set is

.

3.9. Example

Find the solutions of in, and the number of the solutions.

1) If p = 3 and in.

2) If p = 2 and in.

Solution:

1) Since, a = 2, r = 4, p does not divide a, where, m = 1.

So a and r are even, and p is odd. Then by (3.1) the solutions of in are and, so the number of solutions is permutations.

2) Since, a = 3, r = 2, p does not divide a, where, m = 1.

Also, since a is odd and p is even. Then by (3.1) the solution set of in is. So the number of solutions is permutations.

3.10. Remark

If conjugate class of in belong to the solution set of class equation in An and, then we denote to this set by or

.

3.11. Example

Find the solution of

in and the number of the solutions.

Solution:

Let and, since

Then by (3.7) the solutions of

in are, and. So the number of the solutions set is permutations.

4. Concluding Remarks

By the Cayley’s theorem: Every finite group G is isomorphic to a subgroup of the symmetric group, for some. Then we can discuss these propositions. Let be class equation in finite group G and assume that, for some and. The first question we are concerned with is: What is the possible value of d provided that there is no solution for in G? The second question we are concerned with is: what is the possible value of d provided that there is a solution for in G? And then we can find the solution and the number of the solution for in G by using Cayley’s theorem and our theorems in this paper. In another direction, let G be a finite group, and the least positive integer number satisfy. If, then we write

and. For each

and we have. By the Cayley’s theorem we can suppose that or. Also the questions can be summarized as follows:

1) Is collection set of conjugacy classes of G?

2) Is there some, such that for each of, where?

3) Is there some, such that for each of, where?

4) If and is the number of partitions of n, is?

5) If and has m ambivalent conjugacy classes. It is true that is also necessarily G has m ambivalent conjugacy classes?

Finally we will discuss if there is any relation between, and in and what is the possible value of d provided that there is a solution for in G where and for some n to be:

1).

2).

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