The symmetric group 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

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

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 $(\frac{5}{3}) = 10$ 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 $(\frac{5}{3}) = 10$ 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 $(\frac{5}{2})$ ways to select the first two elements; then once we have done so, there are $(\frac{3}{2})$ 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 $(\frac{5}{2}) \times (\frac{3}{2}) / 2 = 15$ elements of this conjugacy class.

The $2, 1, 1, 1$ class

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

There are $(\frac{5}{2})$ 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 $(\frac{5}{2}) = 10$ 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 $1$ .

Conjugacy classes of the symmetric group on five elements