# Math Help - conjugacy classes on symmetric groups

1. ## conjugacy classes on symmetric groups

dear math help forums,

i fear that with the advent of my upcoming abstract algebra exam, the world shall end. however, in order to fend off the end of the world, i shall try my best to defeat this exam, and save the planet. your help in my triumph over this exam will be greatly appreciated!

i'm struggling with the following problem:

Find all the conjugacy classes of elements in the symmetric group S4.

i know that since S4 is not abelian, the conjugacy classes will not be singletons. wikipedia tells me the following:

The symmetric group S4, consisting of all 24 permutations of four elements, has five conjugacy classes, listed with their orders:

• no change (1)
• interchanging two (6)
• a cyclic permutation of three (8)
• a cyclic permutation of all four (6)
• interchanging two, and also the other two (3)

however, the numbers after these statements are mystifying to me. any help in explaining how these numbers came into fruition would be crucial to saving our home Earth.

2. Originally Posted by chrisneedshelp
Find all the conjugacy classes of elements in the symmetric group S4.
Remember if $\sigma \in S_n$ then we can write $\sigma = \theta_1 \theta_2 ... \theta_r$ and $r\geq 1$ where $\theta_j$ are cycles so that $\theta_i$ and $\theta_j$ are disjoint for $i\not = j$. Two elements $\sigma,\tau \in S_n$ are said to have the same structure iff when we write $\sigma = \theta_1 \theta_2 ... \theta_r$ and $\tau = \eta_1 \eta_2 ... \eta_s$ then $r=s$ and can be relabled so that $\theta_j$ has the same cycle length as $\eta_j$. Two elements $\sigma,\tau$ are conjugate to eachother if and only if $\sigma$ and $\tau$ have the same structure.

If you take $\sigma \in S_4$ then we have a certain number of possibilities:
• $\sigma$ is 1-cycle i.e. identity --> there is only one element that has this property
• $\sigma$ is a 2-cycle --> there are ${4\choose 2} = 6$ such elements
• $\sigma$ is a 3-cycle --> there are $2! {4\choose 3} = 8$ such elements
• $\sigma$ is a 4-cycle --> there are $3! = 6$ such elements
• $\sigma$ is a product of two 2-cycles --> there are $\frac{1}{2}{4\choose 2} = 3$ such elements

As a check confirm that $1+3+6+6+8 = 24$ (this happens to be the conjugacy class equation).

3. Originally Posted by ThePerfectHacker
• $\sigma$ is 1-cycle i.e. identity --> there is only one element that has this property
• $\sigma$ is a 2-cycle --> there are ${4\choose 2} = 6$ such elements
• $\sigma$ is a 3-cycle --> there are $2! {4\choose 3} = 8$ such elements
• $\sigma$ is a 4-cycle --> there are $3! = 6$ such elements
• $\sigma$ is a product of two 2-cycles --> there are $\frac{1}{2}{4\choose 2} = 3$ such elements
The 1 cycle identity is just (1,2,3,4) followed by (1,2,3,4).
the 2 cycles would be all permutations of moving elements once? Is this correct? I'm not sure about these values, myself. I've probably missed something along the way that makes these values make sense.

4. Originally Posted by chrisneedshelp
The 1 cycle identity is just (1,2,3,4) followed by (1,2,3,4).
the 2 cycles would be all permutations of moving elements once? Is this correct? I'm not sure about these values, myself. I've probably missed something along the way that makes these values make sense.

$\{ \text{id} \}$
$\{ (12)(34), (13)(24),(14)(23) \}$
$\{ (12),(13),(14),(23),(24),(34) \}$
$\{(123),(124),(132),(134),(142),(143),(234),(243)\ }$
$\{ (1234),(1243),(1324),(1343),(1423),(1432)\}$