Conjugacy classes in $S_5$ by cycle type, corresponding to partitions of $5$

This follows because of the result on homework 1, which says that for any two cycles \(\tau,\sigma\in S_n\), we have \[\tau(s_1~s_2~\cdots~s_k)\tau^{-1} = (\tau(s^1)~\tau(s^2)~\cdots~\tau(s_k)).\]

In particular, this shows that the cycle type of a single cycle is invariant under conjugation. If an element \(\sigma \in S_n\) is comprised of multiple cycles, say \(\sigma = \sigma_1 \cdots \sigma_\ell\), then \[\tau(\sigma)\tau^{-1} = \tau(\sigma_1 \cdots \sigma_\ell)\tau^{-1} = (\tau\sigma_1\tau^{-1})\cdots(\tau\sigma_\ell\tau^{-1}),\]

which shows that the entire cycle type is preserved under conjugation. So each conjugacy class has exactly one cycle type, and distinct classes have distinct cycle types, so this completely determines the conjugacy classes.

Claim 2: Cycle types in \(S_n\) are in bijective correspondence with integer partitions of \(n\).

This follows because any integer partition of \(n\) can be used to obtain a canonical representative of a conjugacy class of \(S_n\): if \(n = a_1 + a_2 + \cdots a_n\), we simply take a cycle of length \(a_1\) the first \(a_1\) integers in order, a cycle of length \(a_2\) containing the integers \(a_1 + 1\) to \(a_2\) in order, and so on.

Conversely, any permutation can be written as a product of disjoint cycles, and when the cycles for fixed points are added in, every integer between 1 and \(n\) will appear, and the sum of the lengths of all cycles must sum to \(n\). Thus taking the cycle lengths yields an integer partition of \(n\).

All integer partitions of 5 are given below, along with a canonical representative of the associated conjugacy class. \[\begin{aligned} 5 && (1~2~3~4~5) \\ 4 + 1 && (1~2~3~4)(5) \\ 3 + 2 && (1~2~3)(4~5) \\ 3 + 1 + 1 && (1~2~3)(4)(5) \\ 2 + 2 + 1 && (1~2)(3~4)(5) \\ 2 + 1 + 1 + 1 && (1~2)(3)(4)(5) \\ 1 + 1 + 1 + 1 + 1 && (1)(2)(3)(4)(5) \\\end{aligned}\] ◻