Counting Cycles in S_5: Examining Conjugacy Classes and Lengths

[joeblow]

I am examining the conjugacy classes of S_5. I know that two elements in S_5 are conjugate if and only if they have the same structure when expressed as products of irreducible cycles. Thus, the conjugacy classes are [(1 2 3 4 5)], [(1 2 3 4)], [(1 2 3)], [(1 2 3)(4 5)], [(1 2)(3 4)], [(1 2)], [(1)]. Now, I want to find how many elements of S_5 are in each.

I need to find the number of cycles of length 2,3,4, and 5 in S_5. I know there are C(5,2)=10 cycles of length 2. I was thinking that to count the cycles of length 3, I start with a cycle of length 2, chose 1 of the 3 unused elements and insert it in one of two possible places to put it. (e.g. If I start with the cycle (1 2) and I choose to add 3, I can make (3 1 2) = (1 2 3) or (1 3 2) )

This should give C(10,1) * C(3,1) * 2 = 60 elements in [(1 2 3)]. However, since [(1 2 3)(4 5)] should have the same number of elements as [(1 2 3)] (since the 2-cycle is set once the 3-cycle is chosen), this would give more elements than are possible.

What am I forgetting to subtract off?

 

[micromass]

I am examining the conjugacy classes of S_5. I know that two elements in S_5 are conjugate if and only if they have the same structure when expressed as products of irreducible cycles. Thus, the conjugacy classes are [(1 2 3 4 5)], [(1 2 3 4)], [(1 2 3)], [(1 2 3)(4 5)], [(1 2)(3 4)], [(1 2)], [(1)]. Now, I want to find how many elements of S_5 are in each.

I need to find the number of cycles of length 2,3,4, and 5 in S_5. I know there are C(5,2)=10 cycles of length 2. I was thinking that to count the cycles of length 3, I start with a cycle of length 2, chose 1 of the 3 unused elements and insert it in one of two possible places to put it. (e.g. If I start with the cycle (1 2) and I choose to add 3, I can make (3 1 2) = (1 2 3) or (1 3 2) )

This should give C(10,1) * C(3,1) * 2 = 60 elements in [(1 2 3)]. However, since [(1 2 3)(4 5)] should have the same number of elements as [(1 2 3)] (since the 2-cycle is set once the 3-cycle is chosen), this would give more elements than are possible.

What am I forgetting to subtract off?

You count many elements many times.

For example, you can start with (1 2) and add 3 to make (1 2 3) or (1 3 2).
But you could also start with (2 3) and add 1 to make (1 2 3) or (1 3 2).

So you at least count every element twice, and perhaps even more.