Conjugacy Classes of Symmetric Group

Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Let $S_n$ denote the symmetric group on $n$ letters.

The conjugacy classes of $S_n$ are determined entirely by the cycle type.

That is, the conjugacy class $\conjclass x$ of an element $x$ of $S_n$ consists of all the elements of $S_n$ whose cycle type is the same as the cycle type of $x$.

Proof
Let $\sigma \in S_n$ have cycle type $\tuple {k_1, k_2, \ldots, k_n}$.

Let $\rho$ be conjugate to $\sigma$

From Conjugate Permutations have Same Cycle Type, $\rho$ has the same cycle type $\tuple {k_1, k_2, \ldots, k_n}$ as $\sigma$.

That is, all the elements of the same conjugacy class have the same cycle type.

Let $\sigma, \rho \in S_n$ have the same cycle type $\tuple {k_1, k_2, \ldots, k_n}$.

It is to be demonstrated that $\sigma$ and $\rho$ are in the same conjugacy class.

From Existence and Uniqueness of Cycle Decomposition, $\sigma$ and $\rho$ can each be expressed uniquely as the product of disjoint cycles:

where $\alpha_i$ and $\beta_i$ are $k_i$-cycles.

For each $i$, let the $k_i$-cycles $\alpha_i$ and $\beta_i$ be expressed as:

For all $i, j$ such that $1 \le i \le l$, $1 \le j \le k_i$, let:
 * $\tau := \map \tau {\alpha_{i j} } = \beta_{i j}$

Such a $\tau$ is bound to exist in $S_n$, as the underlying set of $S_n$ is the set of all permutations of $\set {1, 2, \ldots, n}$.

Thus from Product of Conjugates equals Conjugate of Products:
 * $\tau \alpha_i \tau^{-1} = \beta_i$

Hence:

demonstrating that $\sigma$ and $\rho$ are conjugate.

That is, $\sigma$ and $\rho$ are in the same conjugacy class.

Hence the result.