# 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.