# 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.

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