Conjugacy classes of a symmetric group are determined by cycle structure

Two permutations $𝜎 1, 𝜎 2 \in 𝑆 𝑛$ are conjugate iff they have the same number of $𝑗$ -cycles $𝑘 𝑗 (𝜎 1) = 𝑘 𝑗 (𝜎 2)$ for each $𝑗 = 1, \dots, 𝑛$ . #m/thm/group

Proof

Let $𝜎, 𝜏 \in 𝑆 𝑛$ where $𝜎$ the product of $𝑚$ disjoint cycles

𝜎 = 𝛼 1 𝛼 2 \dots 𝛼 𝑚 - 1 𝛼 𝑚

Then conjugating $𝜎$ by $𝜏$ is the same as the product of conjugating each cycle

𝜏 𝜎 𝜏 - 1 = 𝜏 𝛼 1 𝜏 - 1 𝜏 𝛼 2 𝜏 - 1 \dots 𝜏 𝛼 𝑚 - 1 𝜏 - 1 𝜏 𝛼 𝑚 𝜏 - 1

but The conjugate of an $𝑛$ -cycle is an $𝑛$ -cycle, hence the cycle structure of $𝜏 𝛼 𝜏 - 1$ is identical.

The conjugacy classes of $𝑆 𝑛$ thus correspond to partitions of $𝑛$ .

#state/tidy | #lang/en | #SemBr