# Is there a fast way to determine the conjugacy classes of $S_5$?

I was working towards proving $A_5$ is the only nontrivial normal subgroup of $S_5$. To do this, I wanted to find a set of representatives of conjugacy classes of $S_5$, and their respective orders.

With some research, I found that $1, (12), (123), (12)(34), (1234), (12)(345), (12345)$ are a set of representatives, with orders $1,10,20,15,30,20,24$. What is the most efficient way to determine something like this by hand, without taking an element $\sigma$, conjugating it by all $120$ elements of $S_5$, and then doing it again with a new element which wasn't in the orbit of $\sigma$ until the set is exhausted?

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Conjugacy classes in $S_n$ are just the cycle types. –  Dylan Moreland Jun 1 '12 at 6:54
–  Martin Sleziak Jun 1 '12 at 7:11
Thanks for the info. –  Connor Gillaspie Jun 1 '12 at 7:13

Consider the effect of conjugation on cycles. Given the cycle $c=(a_1~a_2~\cdots~a_k)$ and $\sigma\in S_n$, the permutation $\sigma c\sigma^{-1}$ maps $\sigma(a_i)$ to $\sigma(a_{i+1})$ (cycle the index modulo $k$), and all other arguments are fixed by the permutation. Hence

$$c^\sigma=\sigma(a_1~a_2~\cdots~a_k)\sigma^{-1}=(\sigma (a_1)~\sigma(a_2)~\cdots~\sigma(a_k)).$$

If $c_1,\cdots,c_l$ are disjoint cycles, $\sigma(c_1\cdots c_l)\sigma^{-1}=(\sigma c_1\sigma^{-1})\cdots(\sigma c_l\sigma^{-1})$, hence conjugation preserves not only cycles and cycle lengths, but cycle types as well. Moreover, conjugation acts transitively on permutations of a given cycle type. Suppose $\pi=c_1\cdots c_l$ and $\rho=d_1\cdots d_l$ are two permutations and their disjoint cycle decompositions such that each $c_i$ and $d_i$ have the same length.

We construct a $\sigma$ in pieces: if $c_i=(a_1~a_2~\cdots~a_k)$ and $d_i=(b_1~b_2~\cdots~b_k)$ (note $k$ can vary with $i$), let $\sigma$ send $a_j\mapsto b_j$ for $j=1,2,\cdots,k$. After doing this for each cycle (even the ones of length one, which are normally left out of written representations for efficiency), we will have that $\sigma \pi \sigma^{-1}=\rho$, by having equivalent cycle decompositions. This $\sigma$ is not unique; we could rewrite $c_i$ cycled by $1$ in index as $(a_k~a_1~\cdots~a_{k-1})$ and $\sigma$ would turn out different.

Cycle types are in bijection with integer partitions. For $n=5$, we have

$$\begin{array}{c | c}\lambda\vdash5 & \sigma \\ \hline (5) & (1~2~3~4~5) \\ (4,1) & (1~2~3~4)(5) \\ (3,2) & (1~2~3)(4~5) \\ (3,1,1) & (1~2~3)(4)(5) \\ (2,2,1) & (1~2)(3~4)(5) \\ (2,1,1,1) & (1~2)(3)(4)(5) \\ (1,1,1,1,1) & (1)(2)(3)(4)(5) \end{array}$$

On the left are integer partitions of $5$, and on the right are easily constructed representatives (in cycle notation) of the conjugacy class - cycle type - associated to the partition.

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If $\pi$ and $\sigma$ are permutations, then $\sigma\pi\sigma^{-1}$ has the same cycle type as $\pi$. To see why this is, suppose $\pi$ contains the cycle $(12345)$. Then $\sigma\pi\sigma^{-1}$ takes $\sigma(1)$ to $1$, then to $2$ and then to $\sigma(2)$. So the cycle $(12345)$ turns into the cycle $(\sigma(1)\; \sigma(2)\; \sigma(3)\; \sigma(4)\; \sigma(5) )$.

This means that the conjugacy classes in $S_n$ are exactly the sets of permutations which have the same cycle type when written in disjoint cycle notation. In $A_n$, the problem is a little more complicated, as two permutations $\pi_1$ and $\pi_2$ that have the same cycle type might not have an even permutation $\sigma$ such that $\pi_2=\sigma\pi_1\sigma^{-1}$. It turns out that the conjugacy classes in $A_n$ are the sets of permutations with the same cycle type, unless that cycle type consists of odd cycles of distinct lengths, in which case there are two conjugacy classes of permutations with that cycle type.

I hope you can now see that the representative elements you found for each of the conjugacy classes each have a different cycle type and that the permutations in each conjugacy class in $S_n$ are exactly those permutations with the same cycle type as each representative element.

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I am just going to throw this out there. I hope it helps. I am new here so I don't know how to type things to make them look nice like in LaTex. I know that there is a theorem that states that all $A_n$, where $n$ is greater than or equal to $5$ are simple. A group is simple if it's only normal subgroups are the identity subgroup or itself. So, $A_5$ is simple. Meaning that it is a normal subgroup of $S_5$.
That $A_5$ is a normal subgroup of $S_5$ has pretty much nothing to do with $A_5$ being simple. After all, $A_4$ is not simple, but it's a normal subgroup of $S_4$. –  Gerry Myerson Jun 1 '12 at 7:34