# Conjugacy classes of the alternating group on five elements: Simpler proof

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.

Parents: