Conjugacy classes of the symmetric group on five elements

The sym­met­ric group \(S_5\) on five gen­er­a­tors has size \(5! = 120\) (where the ex­cla­ma­tion mark de­notes the fac­to­rial func­tion). By the re­sult that in a sym­met­ric group, con­ju­gacy classes co­in­cide with cy­cle types, we can list the con­ju­gacy classes of \(S_5\) eas­ily.

The table

$$\begin{array}{|c|c|c|c|} \hline \text{Representative}& \text{Size of class} & \text{Cycle type} & \text{Order of element} \\ \hline (12345) & 24 & 5 & 5 \\ \hline (1234) & 30 & 4,1 & 4 \\ \hline (123) & 20 & 3,1,1 & 3 \\ \hline (123)(45) & 20 & 3,2 & 6 \\ \hline (12)(34) & 15 & 2,2,1 & 2 \\ \hline (12) & 10 & 2,1,1,1 & 2 \\ \hline e & 1 & 1,1,1,1,1 & 1 \\ \hline \end{array}$$

Deter­min­ing the list of cy­cle types and sizes

There are five el­e­ments to per­mute; we need to find all the pos­si­ble ways of group­ing them up into dis­joint cy­cles. We will go through this sys­tem­at­i­cally. Note first that since there are only five el­e­ments to per­mute, there can­not be any el­e­ment with a \(6\)-cy­cle or higher.

Those with largest cy­cle of length \(5\)

If there is a \(5\)-cy­cle, then it per­mutes ev­ery­thing, so its cy­cle type is \(5\). That is, we can take a rep­re­sen­ta­tive \((12345)\), and this is the only con­ju­gacy class with a \(5\)-cy­cle.

Re­call that ev­ery cy­cle of length \(5\) may be writ­ten in five differ­ent ways: \((12345)\) or \((23451)\) or \((34512)\), and so on. Without loss of gen­er­al­ity, we may as­sume \(1\) comes at the start (by cy­cling round the per­mu­ta­tion if nec­es­sary).

Then there are \(4!\) ways to fill in the re­main­ing slots of the cy­cle (where the ex­cla­ma­tion mark de­notes the fac­to­rial func­tion).

Hence there are \(24\) el­e­ments of this con­ju­gacy class.

Those with largest cy­cle of length \(4\)

If there is a \(4\)-cy­cle, then it per­mutes ev­ery­thing ex­cept one el­e­ment, so its cy­cle type must be \(4,1\) (ab­bre­vi­ated as \(4\)). That is, we can take a rep­re­sen­ta­tive \((1234)\), and this is the only con­ju­gacy class with a \(4\)-cy­cle.

Either the el­e­ment \(1\) is per­muted by the \(4\)-cy­cle, or it is not.

  • In the first case, we have \(4\) pos­si­ble ways to pick the image \(a\) of \(1\); then \(3\) pos­si­ble ways to pick the image \(b\) of \(a\); then \(2\) pos­si­ble ways to pick the image \(c\) of \(b\); then \(c\) must be sent back to \(1\). That is, there are \(4 \times 3 \times 2 = 24\) pos­si­ble \(4\)-cy­cles con­tain­ing \(1\).

  • In the sec­ond case, the cy­cle does not con­tain \(1\), so there are \(3\) pos­si­ble ways to pick the image \(a\) of \(2\); then \(2\) pos­si­ble ways to pick the image \(b\) of \(a\); then \(1\) pos­si­ble way to pick the image \(c\) of \(b\); then \(c\) must be sent back to \(2\). So there are \(3 \times 2 \times 1 = 6\) pos­si­ble \(4\)-cy­cles not con­tain­ing \(1\).

This comes to a to­tal of \(30\) pos­si­ble \(4\)-cy­cles in this con­ju­gacy class.

Those with largest cy­cle of length \(3\)

Now we have two pos­si­ble con­ju­gacy classes: \(3,1,1\) and \(3,2\).

The \(3,1,1\) class

An ex­am­ple rep­re­sen­ta­tive for this class is \((123)\).

We pro­ceed with a slightly differ­ent ap­proach to the \(4,1\) case. Us­ing the no­ta­tion for the bino­mial co­effi­cient, we have \(\binom{5}{3} = 10\) pos­si­ble ways to se­lect the num­bers which ap­pear in the \(3\)-cy­cle. Each se­lec­tion has two dis­tinct ways it could ap­pear as a \(3\)-cy­cle: the se­lec­tion \(\{1,2,3\}\) can ap­pear as \((123)\) (or the du­pli­cate cy­cles \((231)\) and \((312)\)), or as \((132)\) (or the du­pli­cate cy­cles \((321)\) or \((213)\)).

That is, we have \(2 \times 10 = 20\) el­e­ments of this con­ju­gacy class.

The \(3,2\) class

An ex­am­ple rep­re­sen­ta­tive for this class is \((123)(45)\).

Again, there are \(\binom{5}{3} = 10\) pos­si­ble ways to se­lect the num­bers which ap­pear in the \(3\)-cy­cle; hav­ing made this se­lec­tion, we have no fur­ther choice about the \(2\)-cy­cle.

Given a se­lec­tion of the el­e­ments of the \(3\)-cy­cle, as be­fore we have two pos­si­ble ways to turn it into a \(3\)-cy­cle.

We are also given the se­lec­tion of the el­e­ments of the \(2\)-cy­cle, but there is no choice about how to turn this into a \(2\)-cy­cle be­cause, for in­stance, \((12)\) is equal to \((21)\) as cy­cles. So this time the se­lec­tion of the el­e­ments of the \(3\)-cy­cle has au­to­mat­i­cally de­ter­mined the cor­re­spond­ing \(2\)-cy­cle.

Hence there are again \(2 \times 10 = 20\) el­e­ments of this con­ju­gacy class.

Those with largest cy­cle of length \(2\)

There are two pos­si­ble cy­cle types: \(2,2,1\) and \(2,1,1,1\).

The \(2,2,1\) class

An ex­am­ple rep­re­sen­ta­tive for this class is \((12)(34)\).

There are \(\binom{5}{2}\) ways to se­lect the first two el­e­ments; then once we have done so, there are \(\binom{3}{2}\) ways to se­lect the sec­ond two. Hav­ing se­lected the el­e­ments moved by the first \(2\)-cy­cle, there is only one dis­tinct way to make them into a \(2\)-cy­cle, since (for ex­am­ple) \((12)\) is equal to \((21)\) as per­mu­ta­tions; similarly the se­lec­tion of the el­e­ments de­ter­mines the sec­ond \(2\)-cy­cle.

How­ever, this time we have dou­ble-counted each el­e­ment, since (for ex­am­ple) the per­mu­ta­tion \((12)(34)\) is equal to \((34)(12)\) by the re­sult that dis­joint cy­cles com­mute.

There­fore there are \(\binom{5}{2} \times \binom{3}{2} / 2 = 15\) el­e­ments of this con­ju­gacy class.

The \(2,1,1,1\) class

An ex­am­ple rep­re­sen­ta­tive for this class is \((12)\).

There are \(\binom{5}{2}\) ways to se­lect the two el­e­ments which this cy­cle per­mutes. Once we have se­lected the el­e­ments, there is only one dis­tinct way to put them into a \(2\)-cy­cle, since (for ex­am­ple) \((12)\) is equal to \((21)\) as per­mu­ta­tions.

There­fore there are \(\binom{5}{2} = 10\) el­e­ments of this con­ju­gacy class.

Those with largest cy­cle of length \(1\)

There is only the iden­tity in this class, so it is of size \(1\).

Parents:

  • Symmetric group

    The sym­met­ric groups form the fun­da­men­tal link be­tween group the­ory and the no­tion of sym­me­try.