1. ## conjugacy classes

i have to show
if n is odd then there are exactly two conjugacy classes of n cycles in An each of which contains (n-1)!/2 elements.
also there is a hint says let An act on itself
i know the fact that since An acts on itself, x doesnt commute with any odd permutation. So it splits into two Ccl An (x) and CCl An (12)x(12).
First of all i have no idea about how to find the elements of CCl An (12)x(12). For sure there are n!/2 elements of An and it will have the half of the number of elements that Sn does.
the number of the elements in CCl Sn (x) is n (n-1)...(n-k+1)/ k
so we better get k=n!/2
but somehow i cant reach the answer

2. Originally Posted by dreamon
i have to show
if n is odd then there are exactly two conjugacy classes of n cycles in An each of which contains (n-1)!/2 elements.
also there is a hint says let An act on itself
i know the fact that since An acts on itself, x doesnt commute with any odd permutation. So it splits into two Ccl An (x) and CCl An (12)x(12).
First of all i have no idea about how to find the elements of CCl An (12)x(12). For sure there are n!/2 elements of An and it will have the half of the number of elements that Sn does.
the number of the elements in CCl Sn (x) is n (n-1)...(n-k+1)/ k
so we better get k=n!/2
but somehow i cant reach the answer
The easy way: Are you familiar with the orbit-stabiliser theorem? If so, can you think up a way of using this? Which elements commute with your n-cycle?

The harder way: Alternatively, note that when n is odd, conjugation of a long-cycle by (12) must be "undone" by (12). There is no way to get from $(12)\sigma(12)$ without conjugation by (12) when $n$ is odd.

(When dealing with these kinds of problems it is perhaps helpful to remember that $(a \text{ } b \ldots r)^{\sigma} = (a \sigma \text{ } b\sigma \ldots r \sigma)$ - every coefficient is acted on individually by the permutation you are conjugating by).

3. Originally Posted by Swlabr
The easy way: Are you familiar with the orbit-stabiliser theorem? If so, can you think up a way of using this? Which elements commute with your n-cycle?

The harder way: Alternatively, note that when n is odd, conjugation of a long-cycle by (12) must be "undone" by (12). There is no way to get from $(12)\sigma(12)$ without conjugation by (12) when $n$ is odd.

(When dealing with these kinds of problems it is perhaps helpful to remember that $(a \text{ } b \ldots r)^{\sigma} = (a \sigma \text{ } b\sigma \ldots r \sigma)$ - every coefficient is acted on individually by the permutation you are conjugating by).
i know orbit and stabilizer theorem. Since i am working in An all of the permutations are even.
gxg-1=x every element here is even as well.
i feel like after finding something about stabilizers ill use that complementary relationship between orbits and stabilizers. so that i coulf be able to conculude with the relation between conj. classes and orbits. but i dont know what to do about number of stabilizers since i just know n is odd

4. Originally Posted by dreamon
i know orbit and stabilizer theorem. Since i am working in An all of the permutations are even.
gxg-1=x every element here is even as well.
i feel like after finding something about stabilizers ill use that complementary relationship between orbits and stabilizers. so that i coulf be able to conculude with the relation between conj. classes and orbits. but i dont know what to do about number of stabilizers since i just know n is odd
The action you are looking at is conjugation by elements of $A_n$. So, what do we call the orbit of an element when the action is conjugation? What is it's stabiliser called?

In this case, it is easier to find the stabiliser than it is to find the orbit. Do this.

5. Originally Posted by Swlabr
The action you are looking at is conjugation by elements of $A_n$. So, what do we call the orbit of an element when the action is conjugation? What is it's stabiliser called?

In this case, it is easier to find the stabiliser than it is to find the orbit. Do this.
the orbit of x under the conjugation action is the set of conjugates for sure.

gxg-1=x is general action for stabilizers.
Since we are working in An we have n!/2 elements.
So for one x from An we have (n-1)!/2 elements. and since the inverse is unique the stab of it has (n-1)!/2 elements.
but not sure.

6. Originally Posted by dreamon
the orbit of x under the conjugation action is the set of conjugates for sure.

gxg-1=x is general action for stabilizers.
Since we are working in An we have n!/2 elements.
So for one x from An we have (n-1)!/2 elements. and since the inverse is unique the stab of it has (n-1)!/2 elements.
but not sure.
Use the fact that $xgx^{-1} = g$ and the hint I gave you in my first reply to work out what $x$ is. How many choices for $x$ are there?

7. Originally Posted by Swlabr
Use the fact that $xgx^{-1} = g$ and the hint I gave you in my first reply to work out what $x$ is. How many choices for $x$ are there?
how can i explain the difference between when n is odd and when n is even

8. Originally Posted by dreamon
how can i explain the difference between when n is odd and when n is even
Remember, you are in $A_n$...

9. Originally Posted by Swlabr
Remember, you are in $A_n$...
I know An is the group of even permutations but it makes a difference when n is odd or even. I am not talking about the permutations/

10. Originally Posted by dreamon
I know An is the group of even permutations but it makes a difference when n is odd or even. I am not talking about the permutations/
What does the question ask about odd and even? Relate this to the definition of $A_n$.