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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