The Alternating group $~$A_5$~$ on five elements has conjugacy classes very similar to those of the Symmetric group $~$S_5$~$, but where one of the classes has split in two.

Note that $~$A_5$~$ has $~$60$~$ elements, since it is precisely half of the elements of $~$S_5$~$ which has $~$5! = 120$~$ elements (where the exclamation mark is the Factorial function).

Table

$$~$\begin{array}{|c|c|c|c|} \hline \text{Representative}& \text{Size of class} & \text{Cycle type} & \text{Order of element} \\ \hline (12345) & 12 & 5 & 5 \\ \hline (21345) & 12 & 5 & 5 \\ \hline (123) & 20 & 3,1,1 & 3 \\ \hline (12)(34) & 15 & 2,2,1 & 2 \\ \hline e & 1 & 1,1,1,1,1 & 1 \\ \hline \end{array}$~$$

Working

Firstly, the identity is in a class of its own, because $~$\tau e \tau^{-1} = \tau \tau^{-1} = e$~$ for every $~$\tau$~$.

Now, by the same reasoning as in the proof that conjugate elements must have the same cycle type in $~$S_n$~$, that result also holds in $~$A_n$~$.

Hence we just need to see whether any of the cycle types comprise more than one conjugacy class.

Recall that the available cycle types are $~$(5)$~$, $~$(3,1,1)$~$, $~$(2,2,1)$~$, $~$(1,1,1,1,1)$~$ (the last of which is the identity and we have already considered it).

Double-transpositions

All the double-transpositions are conjugate (so the $~$(2,2,1)$~$ cycle type does not split):

• $~$(ab)(cd)$~$ is conjugate to $~$(ab)(ce)$~$ if we conjugate by $~$(ab)(de)$~$; symmetrically this covers all the cases where one of the two transpositions remains the same.
• $~$(ab)(cd)$~$ is conjugate to $~$(ac)(bd)$~$ by $~$(cba)$~$; this covers the case that $~$e$~$ is not introduced.
• $~$(ab)(cd)$~$ is conjugate to $~$(ac)(be)$~$ by $~$(bc)(de)$~$; this covers the remaining cases that $~$e$~$ is introduced and neither of the two transpositions remains the same.

Three-cycles

All the three-cycles are conjugate (so the $~$(3,1,1)$~$ cycle type does not split):

• $~$(abc)$~$ is conjugate to $~$(acb)$~$ by $~$(bc)(de)$~$, so three-cycles are conjugate to their permutations.
• $~$(abc)$~$ is conjugate to $~$(abd)$~$ by $~$(cde)$~$; this covers the case of introducing a single new element to the cycle.
• $~$(abc)$~$ is conjugate to $~$(ade)$~$ by $~$(bd)(ce)$~$; this covers the case of introducing two new elements to the cycle.

Five-cycles

This class does split: I claim that $~$(12345)$~$ and $~$(21345)$~$ are not conjugate. (Once we have this, then the class must split into two chunks, since $~$\{ \rho (12345) \rho^{-1}: \rho \ \text{even} \}$~$ is closed under conjugation in $~$A_5$~$, and $~$\{ \rho (12345) \rho^{-1}: \rho \ \text{odd} \}$~$ is closed under conjugation in $~$A_5$~$. The first is the conjugacy class of $~$(12345)$~$ in $~$A_5$~$; the second is the conjugacy class of $~$(21345) = (12)(12345)(12)^{-1}$~$. The only question here was whether they were separate conjugacy classes or whether their union was the conjugacy class.)

Recall that $~$\tau (12345) \tau^{-1} = (\tau(1), \tau(2), \tau(3), \tau(4), \tau(5))$~$, so we would need a permutation $~$\tau$~$ such that $~$\tau$~$ sends $~$1$~$ to $~$2$~$, $~$2$~$ to $~$1$~$, $~$3$~$ to $~$3$~$, $~$4$~$ to $~$4$~$, and $~$5$~$ to $~$5$~$. The only such permutation is $~$(12)$~$, the transposition, but that is not actually a member of $~$A_5$~$.

Hence in fact $~$(12345)$~$ and $~$(21345)$~$ are not conjugate.