0

How can I intuitively think about why a certain pair of disjoint cycles is a duplicate of another? For example $(12)(34)$ is a duplicate of $(12)(43)$. Is it just a matter of brute forcing all elements of order two and checking which ones are duplicates of each other?

Nobilis
  • 127
  • 1
  • 9
  • 2
    $(12)(34)=(12)(43)$ so it cannot be that LHS is a member of a set and RHS is not.The number of transpositions is even which is enough to conclude that it is an even permutation. – drhab Nov 12 '19 at 09:16
  • 1
    Who told you that $(12)(43)$ isn't in $A_5$? Is it someone who wants to count (or list, or something similar) the number of elements or order $2$ in $A_5$? Because then you want to avoid duplicates. – Arthur Nov 12 '19 at 09:17
  • I suspected that it's a duplicate, if I were to manually enumerate them, would I also need to check for duplicates? – Nobilis Nov 12 '19 at 09:23

1 Answers1

1

In order to count number of elements of order $2$, first you have to be know what should be look of the elements. They should be of even order It can be easily seen elements of order two in $S_n$ have look like $(12)$ and $(12)(34)$, but elements of $(12)$ are not possible. Therefore all the elements of $S_n$ which are of type $(12)(34)$ are the only elements of order $2$ in $A_5$.

$$\text{elements of same cocyle}=\frac{120}{2^2 2!}=\frac{120}{8}=15$$ Therefore total $15$ elements of order $2$ in $A_5.$ For the used formula please see the link Counting cycle structures in $S_n$.

MANI
  • 1,928
  • 1
  • 11
  • 23
  • How can I enumerate those elements without including duplicates? I know the properties they satisfy and how many of them there are, but if you include duplicates, the size of the set is bigger than 15. Is it matter of enumerating all order two permutations and then eliminating the duplicates? – Nobilis Nov 12 '19 at 11:00
  • When you said are not possible, I think you meant are not in $A_n$ – J. W. Tanner Nov 12 '19 at 11:02
  • @J.W.Tanner Yes they can not be inside $A_n$ – MANI Nov 12 '19 at 12:43
  • @Nobilis How are you saying that that the size of set is bigger than $15$? – MANI Nov 12 '19 at 12:44
  • @MANI I'm including the duplicates (in which case it's probably not fair to call it a set) – Nobilis Nov 12 '19 at 13:00
  • I suspect my confusion just stems from how to evaluate duplicates properly. – Nobilis Nov 12 '19 at 13:08
  • @Nobilis That is real a tough job, because for sure you need to count the elements of order 2 of even order. This is for sure they should be of this format $(ab)(cd)$. – MANI Nov 12 '19 at 13:13
  • 1
    @MANI Right, thanks, sounds like I just need to make sure I exclude duplicates when enumerating the set and it seems I need to check manually they're no duplicates of ones already there. I should be able to do it with a bit of combinatorics. – Nobilis Nov 12 '19 at 13:18