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 and
where 
(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)
.