Conjugacy classes of an elements in S_12

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  • #1
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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.
 

Answers and Replies

  • #2
Dick
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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
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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
Dick
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The 1/5! is wrong. You can't interchange a 3 cycle with a 2 cycle.
 
  • #6
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I realize that I can divide by 3 and get the right answer, though I'm sure that's not the right reason.
 
  • #7
Dick
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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
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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
Dick
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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
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I guess I'm not sure how to treat this stuff once I have different types of cycles
 
  • #11
Dick
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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
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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
Dick
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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
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I understand that, I'm just having problems counting them. I haven't taken a combinatorics course or anything.
 
  • #16
Dick
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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?
 

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