I am a new user and I don’t know how to use Latex, so I apologize for my text first. I have two questions.

1: I know that the conjugate of a cycle in $S_n$ will not change its cycle type. And I know the converse is also true, i.e. all the cycle with same cycle type consists of a conjugacy class. But I don’t know how to prove this converse. Is there anyone can help me with this?

2: As above, we can write down all the conjugacy classes of $S_n$ easily since we just need to find all the types of cycle, which exactly corresponds to all the partitions of 1, 2, ..., n. The question is that how we find all the conjugacy classes of $A_n$? The method in dealing with $S_n$ doesn’t work now, since there always are two different cycles with the same cycle type, but they are conjugated by a odd cycle. I think about this for a time, but I can’t find a way to formulate it. So is there someone can help me with this?

Appreciate so much :)


Another detailed proof see: Splitting of conjugacy class in alternating group , as Ethan Bolker quotes.


Let $\sigma$ be an even permutation. If there is an odd permutation commuting with $\sigma$, then $\sigma$ has the same conjugacy class in $A_n$ as in $S_n$. Otherwise the centraliser of $\sigma$ in $S_n$ is a subgroup of $A_n$ and the conjugacy class of $\sigma$ in $S_n$ splits into two classes in $A_n$.

If $\sigma$ has a cycle of even length, then that's an odd permutation centralising $\sigma$. If $\sigma$ has two cycles of the same odd length $k$, then there is an involution swapping these two cycles, and centralising $\sigma$. This involution has $k$ two-cycles and so is an odd permutation.

The only remaining possibility is if the cycle decomposition of $\sigma$ consists of cycles of distinct odd lengths. In this case the centraliser of $\sigma$ is generated by these cycles and so is contained in $A_n$. These are the conjugacy classes of $S_n$ that split into two classes in $A_n$.

  • $\begingroup$ I can follow your proof now, thank you so much. $\endgroup$ – Sam Wong Dec 2 '17 at 19:54

All you have to do is find the conjugacy classes of the following elements and use symmetry:




We know that the cycle representation of $\sigma \tau \sigma^{-1}$ is the same of $\tau$ but changing every element by its image under $\sigma$. /

the conjugacy class of $e$ is clear.

The conjugacy class of $(1,2,3)$ has size $4$ because the centralizer of $(1,2,3)$ is just its own powers.

The conjugacy class of $(1,2)(3,4)$ has size $3$ because the centralizer of $(1,2)(3,4)$ has size $4$ because it is generated by $(1,3)(2,4)$ and $(1,2)(3,4)$.

Therefore the $3$-cycles are split into two and the double $2$-cycles remain a conjugacy class.

  • $\begingroup$ Could you explain more detail about the last two sections? How can we get the size of conjugacy from its centralizer? And how we determine if split the 3-cycles or 2-cycles based on the size of their conjugacy respectively? Thanks. $\endgroup$ – Sam Wong Nov 27 '17 at 18:15
  • $\begingroup$ In general if one has a group action $G\rightarrow X$ then the size of the orbit of $x$ is the order of $G$ divided by the order of the stabilizer of $x$. In this case $G$ acts over itself via conjugation and the stabilizer of $g$ is its centralizer. $\endgroup$ – Jorge Fernández Hidalgo Nov 27 '17 at 18:19
  • $\begingroup$ Yes I know this result, which is the orbit-stabilizer theorem. So you assume the order of $G$ is 12 here? And I don’t know how we determine if split the cycle types yet. Could you explain it? Thanks. $\endgroup$ – Sam Wong Nov 27 '17 at 18:32
  • $\begingroup$ the order of $A_4$ is $12$. $\endgroup$ – Jorge Fernández Hidalgo Nov 27 '17 at 18:34
  • $\begingroup$ Yes. But I wanna know the case for $A_n$. The case for n should be motivated by your case. Thank you so much. $\endgroup$ – Sam Wong Nov 27 '17 at 18:37

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