# 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

# 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:

- Conjugacy classes of the alternating group on five elements
\(A_5\) has easily-characterised conjugacy classes, based on a rather surprising theorem about when conjugacy classes in the symmetric group split.

- Alternating group
The alternating group is the only normal subgroup of the symmetric group (on five or more generators).

- Alternating group