Conjugacy classes of an elements in S_12

In summary, the conversation discusses the correct method for computing the number of elements in the conjugacy class of a given permutation in S_12. The formula for this is derived from the number of ways to pick 6 pairs out of 12 elements, with a correction factor to account for overcounting and undercounting in certain cases. The correct correction factor is 2!3! for overcounting 3 cycles and 2 for undercounting 3 cycles, leading to a final formula of \binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\frac{1}{6!2!3!
  • #1
cmj1988
23
0
I 'cheated' and used GAP to compute these, but I want to make sure I am thinking through these properly. This isn't homework as much as given answers in an example and I'm not sure how they computed them.

S_12

(1,2)(3,4)(5,6)(7,8)(9,10)(11,12) - 10,395
This stems from
[tex]\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}
[/tex][tex]\frac{1}{6!}[/tex]

The 6! gets rid of redundant combinations. Is this right? I don't have a calculator handy so I'm not sure how to calculate that.
 
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  • #2
That's right. It's the number of distinct ways to pick 6 pairs out of 12 elements, right?
 
  • #3
I assume you're asking how many elements are in the conjugacy class of (1,2)(3,4)(5,6)(7,8)(9,10)(11,12) in S_12. In that case, the answer that you posted said that you first chose two elements for the first transposition, then two for the second and so on. You divide by 6! because right now, with just the choosing, you're counting

(1,2)(3,4)(5,6)(9,10)(11,12)(7,8) and (7,8)(1,2)(3,4)(5,6)(9,10)(11,12) as two different elements because the transpositions are in a different order. However since they're disjoint, the order doesn't matter.
 
  • #4
Well my next question is, how do I treat this one:

(1,2)(3,4)(5,6)(7,8,9)(10,11,12)

I wanted to treat it as such:

[tex]\binom{12}{3}\binom{9}{3}\binom{6}{2}\binom{4}{2}\frac{1}{5!}[/tex]

However this came out wrong, I believe it undercounted.
 
  • #5
The 1/5! is wrong. You can't interchange a 3 cycle with a 2 cycle.
 
  • #6
I realize that I can divide by 3 and get the right answer, though I'm sure that's not the right reason.
 
  • #7
cmj1988 said:
I realize that I can divide by 3 and get the right answer, though I'm sure that's not the right reason.

Divide what by 3? What number are you looking for? And you have two 3 cycles whose order doesn't matter and three two cycles whose order doesn't matter. What's the correct factorial factor?
 
  • #8
My rationale for pointing out that dividing by 3 gets the answer follows from this:

gap (3 cycle) gap (3 cycle) gap

But another thing is 3/3!, that yields 3 on the bottom
 
  • #9
cmj1988 said:
My rationale for pointing out that dividing by 3 gets the answer follows from this:

gap (3 cycle) gap (3 cycle) gap

But another thing is 3/3!, that yields 3 on the bottom

That's more than a little hard to understand. You don't just have gaps, the gaps have to made of 2 cycles. You were almost right before. You just had the wrong factorial factor.
 
  • #10
I guess I'm not sure how to treat this stuff once I have different types of cycles
 
  • #11
In the first case, you divided by 6! because you had 6 different cycles of the same length. Now you have 2 cycles of length 3 and 3 cycles of length 2. That means you've overcounted by a factor of 2! on the 3 cycles and 3! on the 2 cycles. What should you divide by?
 
  • #12
I would say 2!3!
 
  • #13
If you're writing these in cycle notation, then be careful that you permute any cycles bigger than two.

When you chose two numbers for a transposition, say 2 and 5, then order didn't matter because (2,5) = 2 -> 5, 5 -> 2 = (5,2). But in a three-(or bigger)-cycle, (2,3,4) [tex] \neq[/tex] (2,4,3). But, like in the two cycle, (2,3,4) = (4,2,3). So just make sure that you count all the different permutations that actually produce different three-cycles.
 
  • #14
mathie.girl said:
If you're writing these in cycle notation, then be careful that you permute any cycles bigger than two.

When you chose two numbers for a transposition, say 2 and 5, then order didn't matter because (2,5) = 2 -> 5, 5 -> 2 = (5,2). But in a three-(or bigger)-cycle, (2,3,4) [tex] \neq[/tex] (2,4,3). But, like in the two cycle, (2,3,4) = (4,2,3). So just make sure that you count all the different permutations that actually produce different three-cycles.

mathie.girl makes a very good point! I was forgetting to consider that.
 
Last edited:
  • #15
I understand that, I'm just having problems counting them. I haven't taken a combinatorics course or anything.
 
  • #16
cmj1988 said:
I understand that, I'm just having problems counting them. I haven't taken a combinatorics course or anything.

Ok. Your 2!3! factor for overcounting is right. But now, as mathie.girl pointed out, if you select a 3 element subset that actually represents two different cycles. That means you've undercounted by a factor of 2 for each 3 cycle. What does that make the correction factor to your product of binomial coefficents?
 

1. What are the conjugacy classes of an element in S12?

The conjugacy classes of an element in S12 are the set of all elements that can be obtained by conjugating the given element with any other element in the group. In other words, they are all the elements that are equivalent to the given element under group operations.

2. How many conjugacy classes are there in S12?

There are 11 conjugacy classes in S12, which can be visualized as 11 different "shapes" or "patterns" of elements within the group. These classes are determined by the cycle structure of the elements, where each class corresponds to a unique cycle type.

3. Can an element in S12 belong to more than one conjugacy class?

No, an element in S12 can only belong to one conjugacy class. Each element is only equivalent to itself under group operations, so it can only be part of one unique class.

4. How can I determine the conjugacy class of a specific element in S12?

To determine the conjugacy class of a specific element in S12, you can use the cycle notation to write out all possible permutations of that element and then compare it to the cycle types of the existing classes. The element will belong to the class with the same cycle type as itself.

5. What is the significance of understanding conjugacy classes in S12?

Understanding conjugacy classes in S12 is important in various mathematical applications, such as group theory and representation theory. It helps in identifying and categorizing elements within the group, which can provide insights into the group's structure and properties.

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